In the previous section we saw some examples in Probability.
• In all the examples that we saw so far, we were able to write the 'exact number of possible outcomes'.
♦ That is., we were able to write an exact number in the denominator.
• We can say that, the 'exact number of possible outcomes' is a finite number.
■ But there are cases in which there will be an 'infinite number of possible items'.
• We will now see such an example:
Example 10
In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?
Solution:
• In a musical chair game, at first, the players will be standing ready to move around a row of seats.
• The seats (which are less in number than the number of players) will be facing opposite sides alternately.
• The person playing the music will be facing away from the players and the seats. When this person starts the music, the players begin to move
• When she stops the music, the players will sit. Those who are lucky will be able to get a seat. The others will be out of the game.
• Then another round begins. This time with some seats removed. The last round consists of two players and one chair
• The person playing the music will be unaware of the happenings because, she is looking the other way. She should not see the moving players.
• All she has to do is follow instructions.
• In our problem, the instructions are these:
♦ Start the music at any convenient time
♦ Stop the music within 2 minutes from the start of the music
• Let us mark this 2 minutes on a number line. It is shown in fig.36.1 below.
♦ Point O indicates the beginning of the music
♦ Point A indicates the 2 minutes
♦ Point B indicates the 1⁄2 minutes
• The music can be stopped at any point between O and A
♦ So each point between O and A is a possible outcome
■ How many points are there in between O and A?
• We have seen that, there will be infinite number of points between any two marks on a number line
• Even if any two marks appear to be close to each other, there will be infinite number of points between them.
♦ The hidden points begin to appear when we 'zoom in'. See fig.22.2.
• So it is clear that, in the denominator will be an unimaginably large number.
• What about the numerator?
• In the numerator, we will be writing 'the number favourable outcomes'
• We want to present the probability of stopping the music within half minutes from O
♦ So each point between O and B is a favourable outcome
■ How many points are there between O and B?
• Obviously, there are infinite points
• So the numerator will also be an unimaginably large number
• Then how do we solve this problem?
■ In this situation, the concept of 'scale' comes to our help. We have seen the details about scales here.
• Let the scale be: '1 unit = 1⁄4 minutes'. Then the number line will be as shown in fig.36.2(a) below
♦ There are 8 divisions between O and A
♦ There are 2 divisions between O and B
• Let the scale be: '1 unit = 1⁄8 minutes'. Then the number line will be as shown in fig.36.2(b)
♦ There are 16 divisions between O and A
♦ There are 4 divisions between O and B
• Let the scale be: '1 unit = 1⁄10 minutes'. Then the number line will be as shown in fig.36.2(c)
♦ There are 20 divisions between O and A
♦ There are 5 divisions between O and B
• We can see that, the number of points between any two marks will depend on the scale
■ So what is the relation between 'the scale' an 'the number of points'?
Let us examine each case:
• In fig.36.2(a), scale is: '1 unit = 1⁄4 minutes'
♦ No. of divisions between O and A = (2 minutes ÷ 1⁄4) = (2 minutes × 4) = 8
♦ No. of divisions between O and B = (1⁄2 minutes ÷ 1⁄4) = (1⁄2 minutes × 4) = 2
• In fig.36.2(b), scale is: '1 unit = 1⁄8 minutes'
♦ No. of divisions between O and A = (2 minutes ÷ 1⁄8) = (2 minutes × 8) = 16
♦ No. of divisions between O and B = (1⁄2 minutes ÷ 1⁄8) = (1⁄2 minutes × 8) = 4
• In fig.36.2(c), scale is: '1 unit = 1⁄10 minutes'
♦ No. of divisions between O and A = (2 minutes ÷ 1⁄10) = (2 minutes × 10) = 20
♦ No. of divisions between O and B = (1⁄2 minutes ÷ 1⁄10) = (1⁄2 minutes × 10) = 5
■ We find that, when the 'denominator in the scale' increases, the 'number of divisions' also increases.
• If the 'denominator in the scale' is very large, we will get a great amount of accuracy.
• If there is enough paper space, we will be able to show even one millionth of a second.
♦ Such high degree of accuracy is required in many scientific and engineering problems
■ But for our problem, we see a very useful information:
• The distances OA and OB are multiplied by the same factors in each case
♦ In fig(a), Both OA and OB are multiplied by 4
♦ In fig(b), Both OA and OB are multiplied by 8
♦ In fig(c), Both OA and OB are multiplied by 10
■ So now let us calculate the probability:
• Probability = Number of favourable outcomes⁄Number of possible outcomes
= Number of points between O and B⁄Number of points between O and A
= (OB × scale factor)⁄(OA × scale factor) = OB⁄OA [∵ scale factor being the same, will cancel out]
= (OB ÷ OA) = (1⁄2 ÷ 2) = (1⁄2 ÷ 2⁄1) = (1⁄2 × 1⁄2) = 1⁄4
Example 11
A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Fig.36.3 below:
• What is the probability that it crashed inside the lake shown in blue colour ?
Solution:
1. The length of the outer rectangle is 9 km and it's width is 4.5 km
• So the area is (9 × 4.5) = 40.5 sq km
• The helicopter can be any where within this 40.5 sq km
• So every point inside the (9 × 4.5) rectangle is a possible outcome
2. We want the 'number of possible outcomes in the denominator'
• So we want the 'number of points inside the (9 × 4.5) rectangle
■ How many points are there inside the (9 × 4.5) rectangle?
Let us try to find out:
3. In fig.36.3(b), the area is divided into (1 km ×1 km) squares.
• The number of squares in the (9 × 4.5) rectangle = (9×4.5)⁄(1×1) = (9×4.5) = 40.5 squares
• If we consider each of these squares as a point, we can write 40.5 in the denominator
4. In the numerator, we will be having the favourable outcomes.
• So each point in the (3 × 2.5) inside rectangle is a favourable outcome
• No. of (1 km ×1 km) squares in the inside rectangle = (3×2.5)⁄(1×1) = (3×4.5) = 7.5 squares
• So we can write 7.5 in the numerator
■ Then the probability = 7.5⁄40.5 = 5⁄27
5. But we feel that a (1 km ×1 km) square is too big to be considered as a 'point'
• We can make it smaller. The square can be (1 m ×1 m) or even (1 mm ×1 mm)
• Then we will get a large number of points. Let us see some examples:
6. The number of (1 m ×1 m) squares in the (9 km × 4.5 km) rectangle
= (9×4.5)⁄(0.001×0.001) = (9×4.5×1000000) = (40.5×1000000)squares
7. The number of (1 m ×1 m) squares in the (3 km × 2.5 km) rectangle
= (3×2.5)⁄(0.001×0.001) = (3×2.5×1000000) = (7.5×1000000) squares
■ Then the probability = (7.5×1000000)⁄(40.5×1000000) = 7.5⁄40.5 = 5⁄27
• So we find that, making the squares smaller will not affect the result
• This is because, similar to what we saw in the previous example, the scale factor will be same for numerator and denominator, and they will cancel each other
■ So P(Helicopter crashed in the lake) = (Area of lake⁄Total area of rectangle)
Example 12
In the fig.36.4 below, M is the centre of the larger circle. A smaller circle is drawn inside the larger circle.
The diameter of the smaller circle is equal to the radius of the larger circle. With out looking, a point is marked inside the larger circle. What is the probability that, the point lies
(i) inside the smaller circle
(ii) outside the smaller circle
Solution:
Part (i):
1. Let r be the radius of the larger circle
• Then diameter of the smaller circle = r
• So radius of the smaller circle = r⁄2
2. Area of the larger circle = πr2.
• Area of the smaller circle = π(r⁄2)2 = [1⁄4 × πr2]
3. So probability of the mark to be inside the smaller circle
= P(Inside small) = Area of smaller circle⁄Area of larger circle = [1⁄4 × πr2] ÷ [πr2] = 1⁄4 = 0.25
Part (ii):
1. We want the probability that the mark is on the outside of smaller circle.
Let us denote it as P(Outside small)
2. If the event 'Outside small' occur, 'Inside small' will not occur
• So 'Inside small' and 'Outside small' are complimentary events
• So 'Outside small' is same as 'Inside small'
• So P(Outside small) is same as P(Inside small)
3. Thus we get: P(Outside small) = P(Inside small) = [1 - P(Inside small)] = [1 - 0.25] = 0.75
Example 13
A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?
Solution:
Part (i):
1. Let the probability that the drawn shirt is acceptable to Jimmy be P(J)
• No. of total outcomes possible = 100
• No. of favourable outcomes = 88
2. So P(J) = 88⁄100 = 0.88
Part (ii):
1. Let the probability that the drawn shirt is acceptable to Sujatha be P(S)
• No. of total outcomes possible = 100
• No. of favourable outcomes = (100-4) = 96 (∵ 96 is the number of shirts which do not have any major defects)
2. So P(S) = 96⁄100 = 0.96
Another method for part (ii):
1. Let 'S' be the event of 'acceptable'
• If 'acceptable' occurs, then obviously, 'not acceptable' will not occur
• So 'acceptable' and 'not acceptable' are complimentary events
• We can denote the event of 'not acceptable' as 'S'
2. Now, P(S) = 4⁄100.
• So P(S) = [1-P(S)] = [1-4⁄100] = 96⁄100 = 0.96
Example 14
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
(i) 8? (ii) 13? (iii) less than or equal to 12?
Solution:
We have seen a similar problem before. See solved example 1.14.
When two dice are rolled simultaneously, the possible outcomes can be written in the form of a table:
• The above 36 are the only possible outcomes.
• As the problem involves cases related to 'sum', it is also tabulated for each outcome. The table can be used for all the 3 cases in the question.
(i) Sum is 8
Analysing each outcome, we find that 5 outcomes (Outcomes 12, 17, 22, 27, and 32) satisfies this condition. So the probability of getting 8 as the sum = 5⁄36
(ii) Sum is 13
Analysing each outcome, we find that 0 outcomes satisfies this condition. So the probability of getting 13 as the sum = 0⁄36 = 0
(iii) Sum is less than or equal to 12
Analysing each outcome, we find that all 36 outcomes satisfies this condition. So the probability of getting a sum less than or equal to 12 = 36⁄36 = 1
Example 15
Five cards- the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
Part (i):
1. The 5 cards are: 10, J, Q, K and A
• So the number of possible outcomes is 5
2. There is only one Q. So number of favourable outcomes = 1
• Thus P(Q) = 1⁄5.
Part (ii) (a):
1. Q card is put aside. Now there are only 4 cards: 10, J, K and A
• So the number of possible outcomes is 4
2. There is only one A. So number of favourable outcomes = 1
• Thus P(A) = 1⁄4.
Part (ii) (b):
1. Q card is put aside. Now there are only 4 cards: 10, J, K and A
• So the number of possible outcomes is 4
2. There is zero Q. So number of favourable outcomes = 0
• Thus P(Q) = 0⁄4 = 0
Example 16
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Solution:
Part (i):
1. Number of possible outcomes = 20
2. Number of favourable outcomes = 4
3. So P(Defective) = 4⁄20 = 2⁄10 = 0.2
Part (ii):
1. Number of bulbs remaining = 19
2. So Number of possible outcomes = 19
3. Number of non-defective bulbs at the beginning = (20-4) = 16
4. Number of non-defective bulbs now = (16-1) = 15
• So Number of favourable outcomes = 15
5. So P(Non-defective) = 15⁄19
Example 17
(i) A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
(ii) If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x
Solution:
Part (i):
1. Number of possible outcomes = 12
2. Number of favourable outcomes = x
3. So P(Black) = x⁄12
Part (ii):
1. Number of possible outcomes = (12+6) = 18
2. Number of favourable outcomes = (x+6)
3. So P(Black) = (x+6)⁄18
4. Given that this probability is double that was previously obtained. So we can write:
(x+6)⁄18 = 2 × x⁄12
⟹ (x+6)⁄18 = x⁄6 ⟹ (x+6) = 3x ⟹ 2x = 6 ⟹ x = 3
This completes our present discussion on probability. In the next chapter, we will see some topics in statistics.
• In all the examples that we saw so far, we were able to write the 'exact number of possible outcomes'.
♦ That is., we were able to write an exact number in the denominator.
• We can say that, the 'exact number of possible outcomes' is a finite number.
■ But there are cases in which there will be an 'infinite number of possible items'.
• We will now see such an example:
Example 10
In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?
Solution:
• In a musical chair game, at first, the players will be standing ready to move around a row of seats.
• The seats (which are less in number than the number of players) will be facing opposite sides alternately.
• The person playing the music will be facing away from the players and the seats. When this person starts the music, the players begin to move
• When she stops the music, the players will sit. Those who are lucky will be able to get a seat. The others will be out of the game.
• Then another round begins. This time with some seats removed. The last round consists of two players and one chair
• The person playing the music will be unaware of the happenings because, she is looking the other way. She should not see the moving players.
• All she has to do is follow instructions.
• In our problem, the instructions are these:
♦ Start the music at any convenient time
♦ Stop the music within 2 minutes from the start of the music
• Let us mark this 2 minutes on a number line. It is shown in fig.36.1 below.
♦ Point O indicates the beginning of the music
♦ Point A indicates the 2 minutes
♦ Point B indicates the 1⁄2 minutes
 |
| Fig.36.1 |
♦ So each point between O and A is a possible outcome
■ How many points are there in between O and A?
• We have seen that, there will be infinite number of points between any two marks on a number line
• Even if any two marks appear to be close to each other, there will be infinite number of points between them.
♦ The hidden points begin to appear when we 'zoom in'. See fig.22.2.
• So it is clear that, in the denominator will be an unimaginably large number.
• What about the numerator?
• In the numerator, we will be writing 'the number favourable outcomes'
• We want to present the probability of stopping the music within half minutes from O
♦ So each point between O and B is a favourable outcome
■ How many points are there between O and B?
• Obviously, there are infinite points
• So the numerator will also be an unimaginably large number
• Then how do we solve this problem?
■ In this situation, the concept of 'scale' comes to our help. We have seen the details about scales here.
• Let the scale be: '1 unit = 1⁄4 minutes'. Then the number line will be as shown in fig.36.2(a) below
♦ There are 8 divisions between O and A
♦ There are 2 divisions between O and B
 |
| Fig.36.2 |
♦ There are 16 divisions between O and A
♦ There are 4 divisions between O and B
• Let the scale be: '1 unit = 1⁄10 minutes'. Then the number line will be as shown in fig.36.2(c)
♦ There are 20 divisions between O and A
♦ There are 5 divisions between O and B
• We can see that, the number of points between any two marks will depend on the scale
■ So what is the relation between 'the scale' an 'the number of points'?
Let us examine each case:
• In fig.36.2(a), scale is: '1 unit = 1⁄4 minutes'
♦ No. of divisions between O and A = (2 minutes ÷ 1⁄4) = (2 minutes × 4) = 8
♦ No. of divisions between O and B = (1⁄2 minutes ÷ 1⁄4) = (1⁄2 minutes × 4) = 2
• In fig.36.2(b), scale is: '1 unit = 1⁄8 minutes'
♦ No. of divisions between O and A = (2 minutes ÷ 1⁄8) = (2 minutes × 8) = 16
♦ No. of divisions between O and B = (1⁄2 minutes ÷ 1⁄8) = (1⁄2 minutes × 8) = 4
• In fig.36.2(c), scale is: '1 unit = 1⁄10 minutes'
♦ No. of divisions between O and A = (2 minutes ÷ 1⁄10) = (2 minutes × 10) = 20
♦ No. of divisions between O and B = (1⁄2 minutes ÷ 1⁄10) = (1⁄2 minutes × 10) = 5
■ We find that, when the 'denominator in the scale' increases, the 'number of divisions' also increases.
• If the 'denominator in the scale' is very large, we will get a great amount of accuracy.
• If there is enough paper space, we will be able to show even one millionth of a second.
♦ Such high degree of accuracy is required in many scientific and engineering problems
■ But for our problem, we see a very useful information:
• The distances OA and OB are multiplied by the same factors in each case
♦ In fig(a), Both OA and OB are multiplied by 4
♦ In fig(b), Both OA and OB are multiplied by 8
♦ In fig(c), Both OA and OB are multiplied by 10
■ So now let us calculate the probability:
• Probability = Number of favourable outcomes⁄Number of possible outcomes
= Number of points between O and B⁄Number of points between O and A
= (OB × scale factor)⁄(OA × scale factor) = OB⁄OA [∵ scale factor being the same, will cancel out]
= (OB ÷ OA) = (1⁄2 ÷ 2) = (1⁄2 ÷ 2⁄1) = (1⁄2 × 1⁄2) = 1⁄4
Example 11
A missing helicopter is reported to have crashed somewhere in the rectangular region shown in Fig.36.3 below:
 |
| Fig.36.3 |
Solution:
1. The length of the outer rectangle is 9 km and it's width is 4.5 km
• So the area is (9 × 4.5) = 40.5 sq km
• The helicopter can be any where within this 40.5 sq km
• So every point inside the (9 × 4.5) rectangle is a possible outcome
2. We want the 'number of possible outcomes in the denominator'
• So we want the 'number of points inside the (9 × 4.5) rectangle
■ How many points are there inside the (9 × 4.5) rectangle?
Let us try to find out:
3. In fig.36.3(b), the area is divided into (1 km ×1 km) squares.
• The number of squares in the (9 × 4.5) rectangle = (9×4.5)⁄(1×1) = (9×4.5) = 40.5 squares
• If we consider each of these squares as a point, we can write 40.5 in the denominator
4. In the numerator, we will be having the favourable outcomes.
• So each point in the (3 × 2.5) inside rectangle is a favourable outcome
• No. of (1 km ×1 km) squares in the inside rectangle = (3×2.5)⁄(1×1) = (3×4.5) = 7.5 squares
• So we can write 7.5 in the numerator
■ Then the probability = 7.5⁄40.5 = 5⁄27
5. But we feel that a (1 km ×1 km) square is too big to be considered as a 'point'
• We can make it smaller. The square can be (1 m ×1 m) or even (1 mm ×1 mm)
• Then we will get a large number of points. Let us see some examples:
6. The number of (1 m ×1 m) squares in the (9 km × 4.5 km) rectangle
= (9×4.5)⁄(0.001×0.001) = (9×4.5×1000000) = (40.5×1000000)squares
7. The number of (1 m ×1 m) squares in the (3 km × 2.5 km) rectangle
= (3×2.5)⁄(0.001×0.001) = (3×2.5×1000000) = (7.5×1000000) squares
■ Then the probability = (7.5×1000000)⁄(40.5×1000000) = 7.5⁄40.5 = 5⁄27
• So we find that, making the squares smaller will not affect the result
• This is because, similar to what we saw in the previous example, the scale factor will be same for numerator and denominator, and they will cancel each other
■ So P(Helicopter crashed in the lake) = (Area of lake⁄Total area of rectangle)
Example 12
In the fig.36.4 below, M is the centre of the larger circle. A smaller circle is drawn inside the larger circle.
 |
| Fig.36.4 |
(i) inside the smaller circle
(ii) outside the smaller circle
Solution:
Part (i):
1. Let r be the radius of the larger circle
• Then diameter of the smaller circle = r
• So radius of the smaller circle = r⁄2
2. Area of the larger circle = πr2.
• Area of the smaller circle = π(r⁄2)2 = [1⁄4 × πr2]
3. So probability of the mark to be inside the smaller circle
= P(Inside small) = Area of smaller circle⁄Area of larger circle = [1⁄4 × πr2] ÷ [πr2] = 1⁄4 = 0.25
Part (ii):
1. We want the probability that the mark is on the outside of smaller circle.
Let us denote it as P(Outside small)
2. If the event 'Outside small' occur, 'Inside small' will not occur
• So 'Inside small' and 'Outside small' are complimentary events
• So 'Outside small' is same as 'Inside small'
• So P(Outside small) is same as P(Inside small)
3. Thus we get: P(Outside small) = P(Inside small) = [1 - P(Inside small)] = [1 - 0.25] = 0.75
Example 13
A carton consists of 100 shirts of which 88 are good, 8 have minor defects and 4 have major defects. Jimmy, a trader, will only accept the shirts which are good, but Sujatha, another trader, will only reject the shirts which have major defects. One shirt is drawn at random from the carton. What is the probability that
(i) it is acceptable to Jimmy?
(ii) it is acceptable to Sujatha?
Solution:
Part (i):
1. Let the probability that the drawn shirt is acceptable to Jimmy be P(J)
• No. of total outcomes possible = 100
• No. of favourable outcomes = 88
2. So P(J) = 88⁄100 = 0.88
Part (ii):
1. Let the probability that the drawn shirt is acceptable to Sujatha be P(S)
• No. of total outcomes possible = 100
• No. of favourable outcomes = (100-4) = 96 (∵ 96 is the number of shirts which do not have any major defects)
2. So P(S) = 96⁄100 = 0.96
Another method for part (ii):
1. Let 'S' be the event of 'acceptable'
• If 'acceptable' occurs, then obviously, 'not acceptable' will not occur
• So 'acceptable' and 'not acceptable' are complimentary events
• We can denote the event of 'not acceptable' as 'S'
2. Now, P(S) = 4⁄100.
• So P(S) = [1-P(S)] = [1-4⁄100] = 96⁄100 = 0.96
Example 14
Two dice, one blue and one grey, are thrown at the same time. Write down all the possible outcomes. What is the probability that the sum of the two numbers appearing on the top of the dice is
(i) 8? (ii) 13? (iii) less than or equal to 12?
Solution:
We have seen a similar problem before. See solved example 1.14.
When two dice are rolled simultaneously, the possible outcomes can be written in the form of a table:
• The above 36 are the only possible outcomes.
• As the problem involves cases related to 'sum', it is also tabulated for each outcome. The table can be used for all the 3 cases in the question.
(i) Sum is 8
Analysing each outcome, we find that 5 outcomes (Outcomes 12, 17, 22, 27, and 32) satisfies this condition. So the probability of getting 8 as the sum = 5⁄36
(ii) Sum is 13
Analysing each outcome, we find that 0 outcomes satisfies this condition. So the probability of getting 13 as the sum = 0⁄36 = 0
(iii) Sum is less than or equal to 12
Analysing each outcome, we find that all 36 outcomes satisfies this condition. So the probability of getting a sum less than or equal to 12 = 36⁄36 = 1
Example 15
Five cards- the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.
(i) What is the probability that the card is the queen?
(ii) If the queen is drawn and put aside, what is the probability that the second card picked up is (a) an ace? (b) a queen?
Solution:
Part (i):
1. The 5 cards are: 10, J, Q, K and A
• So the number of possible outcomes is 5
2. There is only one Q. So number of favourable outcomes = 1
• Thus P(Q) = 1⁄5.
Part (ii) (a):
1. Q card is put aside. Now there are only 4 cards: 10, J, K and A
• So the number of possible outcomes is 4
2. There is only one A. So number of favourable outcomes = 1
• Thus P(A) = 1⁄4.
Part (ii) (b):
1. Q card is put aside. Now there are only 4 cards: 10, J, K and A
• So the number of possible outcomes is 4
2. There is zero Q. So number of favourable outcomes = 0
• Thus P(Q) = 0⁄4 = 0
Example 16
(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective ?
Solution:
Part (i):
1. Number of possible outcomes = 20
2. Number of favourable outcomes = 4
3. So P(Defective) = 4⁄20 = 2⁄10 = 0.2
Part (ii):
1. Number of bulbs remaining = 19
2. So Number of possible outcomes = 19
3. Number of non-defective bulbs at the beginning = (20-4) = 16
4. Number of non-defective bulbs now = (16-1) = 15
• So Number of favourable outcomes = 15
5. So P(Non-defective) = 15⁄19
Example 17
(i) A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?
(ii) If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x
Solution:
Part (i):
1. Number of possible outcomes = 12
2. Number of favourable outcomes = x
3. So P(Black) = x⁄12
Part (ii):
1. Number of possible outcomes = (12+6) = 18
2. Number of favourable outcomes = (x+6)
3. So P(Black) = (x+6)⁄18
4. Given that this probability is double that was previously obtained. So we can write:
(x+6)⁄18 = 2 × x⁄12
⟹ (x+6)⁄18 = x⁄6 ⟹ (x+6) = 3x ⟹ 2x = 6 ⟹ x = 3
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