In the previous section we saw some solved examples on probability. In this section, we will see some more examples.
Solved example 1.13
A jar contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the jar. What is the probability that the ball drawn is (i) White (ii) Red (iii) Black (iv) Not red?
Solution:
As there are more than one ball of a kind, let us name them.
• The 3 red balls will be R1, R2 and R3
• The 5 black balls will be B1, B2, B3, B4 and B5
• The 4 white balls will be W1, W2, W3 and W4
Step 1: Write the possible outcomes:
Outcome 1: The drawn ball is R1
Outcome 2: The drawn ball is R2
Outcome 3: The drawn ball is R3
Outcome 4: The drawn ball is B1
Outcome 5: The drawn ball is B2
Outcome 6: The drawn ball is B3
Outcome 7: The drawn ball is B4
Outcome 8: The drawn ball is B5
Outcome 9: The drawn ball is W1
Outcome 10: The drawn ball is W2
Outcome 11: The drawn ball is W3
Outcome 12: The drawn ball is W4
So there are 12 outcomes. We need not write them all. In fact, once we understand the basics, we need to write only this:
Step 1: There are 12 possible outcomes.
This result can be used for all the 4 cases of the question. So for each of the cases, only Step 2 (which is analysis of each outcome) need to be done.
(i) Drawn ball is white:
There are 4 outcomes (outcomes 9, 10, 11 and 12. This is equal to the number of white balls) that are favourable for this event. So the probability = 4⁄12 = 1⁄3 = 33.33%
(ii) Drawn ball is red:
There are 3 outcomes (outcomes 1, 2, and 3. This is equal to the number of red balls) that are favourable for this event. So the probability = 3⁄12 = 1⁄4 = 25%
(iii) Drawn ball is black:
There are 5 outcomes (outcomes 4, 5, 6, 7 and 8. This is equal to the number of black balls) that are favourable for this event. So the probability = 5⁄12 = 41.67%
(iii) Drawn ball is not red:
There are 9 outcomes (outcomes 4 to 12. This is equal to the total number of balls which are not red) that are favourable for this event. So the probability = 9⁄12 = 3⁄4 = 75%
Solved example 1.14
Two dice are rolled simultaneously. Find the probability of getting
(i) an even number as the sum
(ii) a total of atleast 11
(iii) a doublet [a doublet is the situation in which both the dice show the same number on the top face. See the fifth definition given here]
(iv) a doublet of even number
(v) getting a sum divisible by 5
(vi) getting a multiple of 3 as the sum
(vii) getting a multiple of 2 on one die and a multiple of 3 on the other die
(viii) getting sum ≤ 3
Solution:
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 some cases related to 'sum', it is also tabulated for each outcome. The table can be used for all the 8 cases in the question.
(i) an even number as the sum
Analysing each outcome, we find that 18 outcomes (Outcomes 1, 3, 5, 8, 10, 12, 13, 15, 17, 20, 22, 24, 25, 27, 29, 32, 34, 36) satisfies this condition. So the probability of getting an even number as the sum = 18⁄36 = 1⁄2 = 50%
(ii) a total of at least 11
This condition means that, the sum should not be less than 11. It should be 11 or greater than 11. Three outcomes (outcomes 30, 35 and 36) satisfies this condition. So the probability = 3⁄36 = 1⁄12 = 8.33%
(iii) a doublet
6 outcomes (outcomes 1, 8, 15, 22, 29 and 36) satisfies this condition. So the probability = 6⁄36 = 1⁄6 = 8.33%
(iv) a doublet of even numbers
3 outcomes (outcomes 8, 22, and 36) satisfies this condition. So the probability = 3⁄36 = 1⁄12 = 16.67%
(v) a sum divisible by 5
7 outcomes (outcomes 4, 9, 14, 19, 24, 29 and 34) satisfies this condition. So the probability = 7⁄36 = 19.44%
(vi) a multiple of 3 as the sum
12 outcomes (outcomes 2, 5, 7, 10, 15, 18, 20, 23, 25, 28, 33 and 36) satisfies this condition. So the probability = 12⁄36 = 1⁄3 = 33.33%
(vii) a multiple of 2 on one die and a multiple of 3 on the other die
11 outcomes (outcomes 8, 12, 14, 16, 18, 21, 24, 32, 33, 34, and 36) satisfies this condition. So the probability = 11⁄36 = 1⁄3 = 30.55%
(viii) getting sum ≤ 3
3 outcomes (outcomes 1, 2, and 7) satisfies this condition. So the probability = 3⁄36 = 1⁄12 = 8.33%
Solved example 1.15
One card is drawn from a pack of 52 cards, each card being equally likely to be drawn. Find the probability that the card drawn is: Either red card or king
Solution:
Step 1: We have 52 possible outcomes.
Step 2: Let us analyse each outcome:
If the drawn card is red, then we have an event. If the drawn card is a king, then also, we have an event. So
• we can write 'favourable' towards 26 red cards.
• we can write 'favourable' towards 4 king cards.
But 2 red kings are already marked 'favourable' when we mark the 26 red cards. So the total number of favourable outcomes are 26 + 2 = 28. Thus the probability = 28⁄52 = 7⁄13 = 53.84%
Solved example 1.16
The king, queen and jack of clubs are removed from a standard pack of 52 cards. The pack is well shuffled, and one card is drawn. Find the probability that the drawn card is (i) a heart (ii) a king (iii) an '8' of spades (iv) a club
Solution:
Step 1: 3 cards are removed from the pack, which leaves 52 -3 = 49 cards. So there are 49 possible outcomes. This result can be used for all the 4 cases in the question.
(i) a heart
There are 13 hearts in the pack. Each of them will give a favourable outcome. So the probability = 13⁄49
(ii) a king
There are 3 kings left in the pack. Each of them will give a favourable outcome. So the probability = 3⁄49
(iii) an '8' of spades
This is an unique card. There is only one favourable outcome. So the probability = 1⁄49
(iv) a club
There are 10 clubs left in the pack. Each of them will give a favourable outcome. So the probability = 10⁄49
We have now obtained a basic idea about the topic of probability. We have also seen some solved examples. We now know to calculate 'the probability of a certain event to occur'. The probability that we calculate is a 'theoretical' value. Let us now see how it is related to the 'real world events'.
Solved example 1.13
A jar contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the jar. What is the probability that the ball drawn is (i) White (ii) Red (iii) Black (iv) Not red?
Solution:
As there are more than one ball of a kind, let us name them.
• The 3 red balls will be R1, R2 and R3
• The 5 black balls will be B1, B2, B3, B4 and B5
• The 4 white balls will be W1, W2, W3 and W4
Step 1: Write the possible outcomes:
Outcome 1: The drawn ball is R1
Outcome 2: The drawn ball is R2
Outcome 3: The drawn ball is R3
Outcome 4: The drawn ball is B1
Outcome 5: The drawn ball is B2
Outcome 6: The drawn ball is B3
Outcome 7: The drawn ball is B4
Outcome 8: The drawn ball is B5
Outcome 9: The drawn ball is W1
Outcome 10: The drawn ball is W2
Outcome 11: The drawn ball is W3
Outcome 12: The drawn ball is W4
So there are 12 outcomes. We need not write them all. In fact, once we understand the basics, we need to write only this:
Step 1: There are 12 possible outcomes.
This result can be used for all the 4 cases of the question. So for each of the cases, only Step 2 (which is analysis of each outcome) need to be done.
(i) Drawn ball is white:
There are 4 outcomes (outcomes 9, 10, 11 and 12. This is equal to the number of white balls) that are favourable for this event. So the probability = 4⁄12 = 1⁄3 = 33.33%
(ii) Drawn ball is red:
There are 3 outcomes (outcomes 1, 2, and 3. This is equal to the number of red balls) that are favourable for this event. So the probability = 3⁄12 = 1⁄4 = 25%
(iii) Drawn ball is black:
There are 5 outcomes (outcomes 4, 5, 6, 7 and 8. This is equal to the number of black balls) that are favourable for this event. So the probability = 5⁄12 = 41.67%
(iii) Drawn ball is not red:
There are 9 outcomes (outcomes 4 to 12. This is equal to the total number of balls which are not red) that are favourable for this event. So the probability = 9⁄12 = 3⁄4 = 75%
Solved example 1.14
Two dice are rolled simultaneously. Find the probability of getting
(i) an even number as the sum
(ii) a total of atleast 11
(iii) a doublet [a doublet is the situation in which both the dice show the same number on the top face. See the fifth definition given here]
(iv) a doublet of even number
(v) getting a sum divisible by 5
(vi) getting a multiple of 3 as the sum
(vii) getting a multiple of 2 on one die and a multiple of 3 on the other die
(viii) getting sum ≤ 3
Solution:
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 some cases related to 'sum', it is also tabulated for each outcome. The table can be used for all the 8 cases in the question.
(i) an even number as the sum
Analysing each outcome, we find that 18 outcomes (Outcomes 1, 3, 5, 8, 10, 12, 13, 15, 17, 20, 22, 24, 25, 27, 29, 32, 34, 36) satisfies this condition. So the probability of getting an even number as the sum = 18⁄36 = 1⁄2 = 50%
(ii) a total of at least 11
This condition means that, the sum should not be less than 11. It should be 11 or greater than 11. Three outcomes (outcomes 30, 35 and 36) satisfies this condition. So the probability = 3⁄36 = 1⁄12 = 8.33%
(iii) a doublet
6 outcomes (outcomes 1, 8, 15, 22, 29 and 36) satisfies this condition. So the probability = 6⁄36 = 1⁄6 = 8.33%
(iv) a doublet of even numbers
3 outcomes (outcomes 8, 22, and 36) satisfies this condition. So the probability = 3⁄36 = 1⁄12 = 16.67%
(v) a sum divisible by 5
7 outcomes (outcomes 4, 9, 14, 19, 24, 29 and 34) satisfies this condition. So the probability = 7⁄36 = 19.44%
(vi) a multiple of 3 as the sum
12 outcomes (outcomes 2, 5, 7, 10, 15, 18, 20, 23, 25, 28, 33 and 36) satisfies this condition. So the probability = 12⁄36 = 1⁄3 = 33.33%
(vii) a multiple of 2 on one die and a multiple of 3 on the other die
11 outcomes (outcomes 8, 12, 14, 16, 18, 21, 24, 32, 33, 34, and 36) satisfies this condition. So the probability = 11⁄36 = 1⁄3 = 30.55%
(viii) getting sum ≤ 3
3 outcomes (outcomes 1, 2, and 7) satisfies this condition. So the probability = 3⁄36 = 1⁄12 = 8.33%
Solved example 1.15
One card is drawn from a pack of 52 cards, each card being equally likely to be drawn. Find the probability that the card drawn is: Either red card or king
Solution:
Step 1: We have 52 possible outcomes.
Step 2: Let us analyse each outcome:
If the drawn card is red, then we have an event. If the drawn card is a king, then also, we have an event. So
• we can write 'favourable' towards 26 red cards.
• we can write 'favourable' towards 4 king cards.
But 2 red kings are already marked 'favourable' when we mark the 26 red cards. So the total number of favourable outcomes are 26 + 2 = 28. Thus the probability = 28⁄52 = 7⁄13 = 53.84%
Solved example 1.16
The king, queen and jack of clubs are removed from a standard pack of 52 cards. The pack is well shuffled, and one card is drawn. Find the probability that the drawn card is (i) a heart (ii) a king (iii) an '8' of spades (iv) a club
Solution:
Step 1: 3 cards are removed from the pack, which leaves 52 -3 = 49 cards. So there are 49 possible outcomes. This result can be used for all the 4 cases in the question.
(i) a heart
There are 13 hearts in the pack. Each of them will give a favourable outcome. So the probability = 13⁄49
(ii) a king
There are 3 kings left in the pack. Each of them will give a favourable outcome. So the probability = 3⁄49
(iii) an '8' of spades
This is an unique card. There is only one favourable outcome. So the probability = 1⁄49
(iv) a club
There are 10 clubs left in the pack. Each of them will give a favourable outcome. So the probability = 10⁄49
We have now obtained a basic idea about the topic of probability. We have also seen some solved examples. We now know to calculate 'the probability of a certain event to occur'. The probability that we calculate is a 'theoretical' value. Let us now see how it is related to the 'real world events'.
Let us take the example of drawing a ball from the jar containing 2 red balls and 5 blue balls. (Details here) We found out that, the probability of getting a red ball is 2⁄7 and that for a blue ball is 5⁄7. So the blue ball has greater probability.
Suppose there is a game of bet based on drawing a ball from this jar. The game is between two players A and B. It is played as follows:
• Player A chooses a color: red or blue
• The balls are well mixed and player A draws a ball without looking. If the drawn ball is of his chosen color, he wins.
• If the drawn ball is not of his chosen color, then he looses. The winner will then be B
• Drawing a ball by player A completes the game
So we know how the game is played. If A ask us for advice before the starting of the game, we would certainly advise to select blue. Because it has greater probability.
But there is no guarantee that the drawn ball will be blue. As the balls are mixed well, red ball also has the chance for being drawn, though it's probability is less. It may so happen that A chooses blue, and the ball he draws is red. Then, A looses the game.
Let the game be played again. This time also A chooses blue. The balls are mixed well and one ball is drawn. The drawn ball can be red or blue. The player has no control over the outcome. If the ball is red, A has lost again.
So even though blue has a high probability of 5⁄7, player A may loose the game. Then how do we relate theory to practice? The answer is that, when the number of trials increases, the practical values approaches the theoretical values. This can be illustrated based on the above game:
In the above example, one game is complete when a ball is drawn. The winner is decided immediately when a ball is drawn. Instead of this, let the winner be chosen based on 5 results. That is., the ball is drawn once. If it is blue, player A wins. If it is red, he looses. This result is noted down. This completes one cycle. The ball is placed back and mixed well. A ball is drawn again. The result is noted down. In this way, a total of 5 cycles are completed, and 5 cycles completes one game. The winner is decided based on the results of all the 5 cycles. A possible result is tabulated below:
Based on the results in the above table, we can say that Player A has lost the game. Even though blue has a higher probability, it was drawn only 2 times, while red was drawn 3 times. [It may be noted that this is only a possibility]
Let the game consists of 10 cycles. The results are tabulated below.
This time we see that blue is drawn 7 times while red is drawn only 3 times. So player A is the winner. [Again, this is also only one of the many possibilities]
While playing such games, and doing experiments like tossing coins, rolling dice etc., it is seen that, when the number of cycles increases, the actual results come closer and closer to theoretical results. So, if it is decided to base the winner of the above game on the results of a large number of cycles, Player A will surely be the winner.
Let us take the example of tossing a coin. There are 2 players A and B. The coin is tossed once. If it is heads A wins. If it is tails, then B wins. Both A and B have equal chance of winning. Because the probability for both Heads and Tails is 1⁄2. But there can be only one winner. The winner is decided immediately after tossing the coin once.
Instead of this, let there be a large number of cycles. Each cycle consists of tossing the coin once. The winner in each cycle is noted down. After completing all the cycles, the player who has obtained the largest number of wins will be declared the final winner. The tabulation can be done as shown below:
If the number of cycles in the above table is large, we will find that the wins obtained by player A and player B will almost be the same.
The relation between theoretical and practical results is that, when the number of cycles increases, the actual results become closer and closer to theoretical results.
We have completed the present discussion on the basics of Probability. The next level discussion is given in chapter 28. In the next chapter, we will discuss about Graphs.
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