In the previous section we saw some examples and learned some new terms in Probability. In this section, we will see a few more examples.
Example 4
One card is drawn from a well shuffled deck of 52 cards. Calculate the probability that the card will
(i) be an ace (ii) not be an ace
Solution:
We have already seen the probabilities related to the deck of cards in an earlier chapter. There we saw the various features of a deck of cards. (Details here)
For our present problem, we can straight away begin to write the steps:
Part (i):
Step 2: Analyse each outcome
• The requirement given is: The card should be an 'ace'.
• There are four aces. So we can write 'favourable' towards four outcomes.
♦ If we get any one of those four, we have an event. So the numerator is 4
Thus P(A) = 4⁄52 = 1⁄13
Part (ii):
• There are four aces. So we can write 'favourable' towards (52-4) = 48 outcomes.
♦ If we get any one of those 48, we have an event. So the numerator is 48
Thus P(B) = 48⁄52 = 12⁄13.
Easy method for part (ii):
1. Event A is the complementary event of event B
• That is., P(A) = P(complement of B)
2. But complementary event of B is (B)
• So we get: P(A) = P(B)
3. In part (ii), we are trying to find P(B)
• We know that P(B) = 1- P(B)
4. So using the result in step (2) above, we can write:
P(B) = 1 - P(A)
• But in part (i) we obtained P(A) as 1⁄13
• Thus we get P(B) = (1 - 1⁄13) = (12⁄13)
Example 5
Two players Sangeeta and Reshma play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
Solution:
1. Let 'the probability of Sangeeta winning the match' be denoted as P(S)
• So we can write: P(S) = 0.62
2. Let 'the probability of Reshma winning the match' be denoted as P(R)
• We want to find this P(S)
3. The 'event of Reshma winning the match' is same as the 'event of Sangeeta not winning the match'
• In other words, R and S are complimentary
• That is., R is same as (not S)
♦ 'not S' is denoted as 'S'
4. So we can write: R is same as S.
• So P(R) = P(S)
5. We can easily find P(S) because, P(S) is given
• We get: P(S) = (1-P(S)) = (1-0.62) = 0.38
• So from (4), we get: P(R) = P(S) = 0.38
Example 6
Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Solution:
Part (i):
1. Let Savita's birthday be 'k'
• 'k' can be any value like JAN 12, MAR 25, SEP 14 etc.,
• What ever be the value, it is unique. Because, each day among the 365 days in an year is a unique day
• We want to know the probability of Hamida's birthday to be 'not on k'
2. Let us write the outcomes:
Outcome 1: Hamida's birthday is on JAN 1
Outcome 2: Hamida's birthday is on JAN 2
Outcome 3: Hamida's birthday is on JAN 3
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
Outcome 365: Hamida's birthday is on DEC 31
• No outcomes other than the above 365 can possibly occur
• We say that: The number of all possible outcomes is 365
• So in the denominator, we will have 365
3. We want to present the probability of Hamida's birthday to be 'not on k'
• We will denote it as P(Not same)
• If Hamida's birthday is 'not on k', we have an event
4. Towards all the 365 outcomes in step(2), we will be writing 'favourable' except one.
• We will be writing 'not favourable' towards the outcome in which 'Hamida's birthday is on 'k''
5. So we will be having 364 favourable outcomes.
• Thus, in the numerator, we will be having 364
• So P(Not same) = 364⁄365
Part (ii):
• For part (ii) we can use the easy method
1. We want to present the probability of Hamida's birthday to be 'on k'
• We will denote it as P(Same)
2. If Hamida's birthday is on 'k',
• It will not be 'Not same'
• That is., if 'Not same' occur, 'Same' will not occur
♦ The condition of 'Same' not occurring is denoted as 'Same'
• Thus we can write: 'Not same' occurring is equivalent to 'Same' occurring.
• So P(Not same) = P(Same)
3. We want P(Same). For that we can use the general relation: P(E) = 1 - P(E)
• So P(Same) = 1 - P(Same)
• Substituting for P(Same) from step (2), we get:
P(Same) = 1 - P(Not same)
4. From part(i), we have P(Not same) = 364⁄365
• Thus we get: P(Same) = (1 - 364⁄365)= 1⁄365
Example 7
There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?
Solution:
Part (i):
1. Let the girls be denoted as: G1, G2, G3, . . . G25
Let the boys be denoted as: B1, B2, B3, . . . B15
2. Let us write the outcomes:
Outcome 1: The drawn card is G1
Outcome 2: The drawn card is G2
Outcome 3: The drawn card is G3
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
Let the boys be denoted as: B1, B2, B3, . . . B15
2. Let us write the outcomes:
Outcome 1: The drawn card is G1
Outcome 2: The drawn card is G2
Outcome 3: The drawn card is G3
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
• This completes part(ii) of the problem. Note that, once we understand the basics, we need not write the detailed steps. The following steps will be sufficient:
Easy method for Part (ii):
In this section, we saw 3 examples and learned some new terms related to Probability. In the next section, we will see more examples.
Example 4
One card is drawn from a well shuffled deck of 52 cards. Calculate the probability that the card will
(i) be an ace (ii) not be an ace
Solution:
We have already seen the probabilities related to the deck of cards in an earlier chapter. There we saw the various features of a deck of cards. (Details here)
For our present problem, we can straight away begin to write the steps:
Part (i):
• Let P(A) be the probability of 'getting an ace'
Step 1: Write the outcomes
There are 52 possible outcomes. So the denominator is 52Step 1: Write the outcomes
Step 2: Analyse each outcome
• The requirement given is: The card should be an 'ace'.
• There are four aces. So we can write 'favourable' towards four outcomes.
♦ If we get any one of those four, we have an event. So the numerator is 4
Thus P(A) = 4⁄52 = 1⁄13
Part (ii):
• Let P(B) be the probability of 'not getting an ace'
Step 1: Write the outcomes
There are 52 possible outcomes. So the denominator is 52Step 1: Write the outcomes
Step 2: Analyse each outcome
• The requirement given is: The card should not be an 'ace'.• There are four aces. So we can write 'favourable' towards (52-4) = 48 outcomes.
♦ If we get any one of those 48, we have an event. So the numerator is 48
Thus P(B) = 48⁄52 = 12⁄13.
Easy method for part (ii):
1. Event A is the complementary event of event B
• That is., P(A) = P(complement of B)
2. But complementary event of B is (B)
• So we get: P(A) = P(B)
3. In part (ii), we are trying to find P(B)
• We know that P(B) = 1- P(B)
4. So using the result in step (2) above, we can write:
P(B) = 1 - P(A)
• But in part (i) we obtained P(A) as 1⁄13
• Thus we get P(B) = (1 - 1⁄13) = (12⁄13)
Example 5
Two players Sangeeta and Reshma play a tennis match. It is known that the probability of Sangeeta winning the match is 0.62. What is the probability of Reshma winning the match?
Solution:
1. Let 'the probability of Sangeeta winning the match' be denoted as P(S)
• So we can write: P(S) = 0.62
2. Let 'the probability of Reshma winning the match' be denoted as P(R)
• We want to find this P(S)
3. The 'event of Reshma winning the match' is same as the 'event of Sangeeta not winning the match'
• In other words, R and S are complimentary
• That is., R is same as (not S)
♦ 'not S' is denoted as 'S'
4. So we can write: R is same as S.
• So P(R) = P(S)
5. We can easily find P(S) because, P(S) is given
• We get: P(S) = (1-P(S)) = (1-0.62) = 0.38
• So from (4), we get: P(R) = P(S) = 0.38
Example 6
Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Solution:
Part (i):
1. Let Savita's birthday be 'k'
• 'k' can be any value like JAN 12, MAR 25, SEP 14 etc.,
• What ever be the value, it is unique. Because, each day among the 365 days in an year is a unique day
• We want to know the probability of Hamida's birthday to be 'not on k'
2. Let us write the outcomes:
Outcome 1: Hamida's birthday is on JAN 1
Outcome 2: Hamida's birthday is on JAN 2
Outcome 3: Hamida's birthday is on JAN 3
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
Outcome 365: Hamida's birthday is on DEC 31
• No outcomes other than the above 365 can possibly occur
• We say that: The number of all possible outcomes is 365
• So in the denominator, we will have 365
3. We want to present the probability of Hamida's birthday to be 'not on k'
• We will denote it as P(Not same)
• If Hamida's birthday is 'not on k', we have an event
4. Towards all the 365 outcomes in step(2), we will be writing 'favourable' except one.
• We will be writing 'not favourable' towards the outcome in which 'Hamida's birthday is on 'k''
5. So we will be having 364 favourable outcomes.
• Thus, in the numerator, we will be having 364
• So P(Not same) = 364⁄365
Part (ii):
• For part (ii) we can use the easy method
1. We want to present the probability of Hamida's birthday to be 'on k'
• We will denote it as P(Same)
2. If Hamida's birthday is on 'k',
• It will not be 'Not same'
• That is., if 'Not same' occur, 'Same' will not occur
♦ The condition of 'Same' not occurring is denoted as 'Same'
• Thus we can write: 'Not same' occurring is equivalent to 'Same' occurring.
• So P(Not same) = P(Same)
3. We want P(Same). For that we can use the general relation: P(E) = 1 - P(E)
• So P(Same) = 1 - P(Same)
• Substituting for P(Same) from step (2), we get:
P(Same) = 1 - P(Not same)
4. From part(i), we have P(Not same) = 364⁄365
• Thus we get: P(Same) = (1 - 364⁄365)= 1⁄365
Example 7
There are 40 students in Class X of a school of whom 25 are girls and 15 are boys. The class teacher has to select one student as a class representative. She writes the name of each student on a separate card, the cards being identical. Then she puts cards in a bag and stirs them thoroughly. She then draws one card from the bag. What is the probability that the name written on the card is the name of (i) a girl? (ii) a boy?
Solution:
Part (i):
1. Let the girls be denoted as: G1, G2, G3, . . . G25
Let the boys be denoted as: B1, B2, B3, . . . B15
2. Let us write the outcomes:
Outcome 1: The drawn card is G1
Outcome 2: The drawn card is G2
Outcome 3: The drawn card is G3
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
Outcome 25: The drawn card is G25
• This completes part(i) of the problem. Note that, once we understand the basics, we need not write the detailed steps. The following steps will be sufficient:
Part (ii):
1. Let the girls be denoted as: G1, G2, G3, . . . G25
Outcome 26: The drawn card is B1
Outcome 27: The drawn card is B2
Outcome 28: The drawn card is B3
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
Outcome 40: The drawn card is B15
• No outcomes other than the above 40 can possibly occur
• We say that: The number of all possible outcomes is 40
• So in the denominator, we will have 40
3. We want to present the probability of 'a girl's name to be drawn'
• We will denote it as P(G)
• If a girl's name is drawn, we have an event
4. Towards the first 25 outcomes in step(2), we will be writing 'favourable'.
• We will be writing 'not favourable' towards the next 15 outcomes
5. So we will be having 25 favourable outcomes.
• Thus, in the numerator, we will be having 25
• So P(G) = 25⁄40 = 5⁄8
• Total no. of outcomes = (25+15) = 40
• No. of favourable outcomes = 25
• So P(G) = Number of outcomes favourable to G⁄Number of all outcomes = 25⁄40 = 5⁄8
Let the boys be denoted as: B1, B2, B3, . . . B15
2. Let us write the outcomes:
Outcome 1: The drawn card is G1
Outcome 2: The drawn card is G2
Outcome 3: The drawn card is G3
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
Outcome 25: The drawn card is G25
Outcome 26: The drawn card is B1
Outcome 26: The drawn card is B1
Outcome 27: The drawn card is B2
Outcome 28: The drawn card is B3
_ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _
Outcome 40: The drawn card is B15
• No outcomes other than the above 40 can possibly occur
• We say that: The number of all possible outcomes is 40
• So in the denominator, we will have 40
3. We want to present the probability of 'a boy's name to be drawn'
• We will denote it as P(B)
• If a boy's name is drawn, we have an event
4. Towards the first 25 outcomes in step(2), we will be writing 'unfavourable'.
• We will be writing 'favourable' towards the next 15 outcomes
5. So we will be having 15 favourable outcomes.
• Thus, in the numerator, we will be having 15
• So P(G) = 15⁄40 = 3⁄8
• Total no. of outcomes = (25+15) = 40
• No. of favourable outcomes = 15
• So P(B) = Number of outcomes favourable to B⁄Number of all outcomes = 15⁄40 = 3⁄8
• For part (ii) we can use the easy method
1. We want to present the probability of 'a boy's name to be drawn'
• We will denote it as P(B)
2. If 'G' occur, 'B' will not occur
♦ The condition of 'B' not occurring is denoted as 'B'
• Thus we can write: 'G' occurring is equivalent to 'B' occuring.
• So P(G) = P(B)
3. We want P(B). For that we can use the general relation: P(E) = 1 - P(E)
• So P(B) = 1 - P(B)
• Substituting for P(B) from step (2), we get:
P(B) = 1 - P(G)
4. From part(i), we have P(G) = 5⁄8
• Thus we get: P(B) = (1 - 5⁄8) = 3⁄8
■ Note that, once we understand the basics, this easy method can also be written in a lesser number of steps:
1. If 'G' occur, 'B' will not occur
♦ The condition of 'B' not occurring is denoted as 'B'
• That is., 'G' occurring is same as 'B' occurring
• So P(G) = P(B)
2. We want P(B), which is equal to [1-P(B)]
• But from step (1), we have P(B) = P(G)
• So P(B) = [1-P(B)] = [1-P(G)] = [1-5⁄8] = 3⁄8
Example 8:
A box contains 3 blue, 2 white, and 4 red marbles. If a marble is drawn at random from the box, what is the probability that it will be (i) white? (ii) blue? (iii) red?
Solution:
Part (i):
1. Let the 3 blue marbles be denoted as: B1, B2 and B3
• Let the 2 white marbles be denoted as: W1 and W2
• Let the 4 red marbles be denoted as: R1, R2, R3 and R4
2. Let us write the outcomes:
Outcome 1: The drawn marble is B1
Outcome 2: The drawn marble is B2
Outcome 3: The drawn marble is B3
Outcome 4: The drawn marble is W1
Outcome 5: The drawn marble is W2
Outcome 6: The drawn marble is R1
Outcome 7: The drawn marble is R2
Outcome 8: The drawn marble is R3
Outcome 9: The drawn marble is R4
• No outcomes other than the above 9 can possibly occur
• We say that: The number of all possible outcomes is 9
• So in the denominator, we will have 9
3. We want to present the probability of 'a white marble to be drawn'
• We will denote it as P(W)
• If a white marble is drawn, we have an event
4. Towards the first 3 outcomes in step(2), we will be writing 'not favourable'.
• Towards the next 2 outcomes in step(2), we will be writing 'favourable'.
• Towards the last 4 outcomes in step(2), we will be writing 'not favourable'.
• Towards the next 2 outcomes in step(2), we will be writing 'favourable'.
• Towards the last 4 outcomes in step(2), we will be writing 'not favourable'.
5. So we will be having 2 favourable outcomes.
• Thus, in the numerator, we will be having 2
• So P(W) = 2⁄9
■ The above steps can be shortened as:
• Total no. of outcomes = (3+2+4) = 9
• No. of favourable outcomes = 2
• So P(W) = Number of outcomes favourable to W⁄Number of all outcomes = 2⁄9
Part (ii):
• Total no. of outcomes = (3+2+4) = 9
• No. of favourable outcomes = 3
• So P(B) = Number of outcomes favourable to B⁄Number of all outcomes = 3⁄9
Part (iii):
• Total no. of outcomes = (3+2+4) = 9
• No. of favourable outcomes = 4
• So P(R) = Number of outcomes favourable to R⁄Number of all outcomes = 4⁄9.
■ Note that P(W) + P(B) + P(R) = [2⁄9 + 3⁄9 + 4⁄9] = [9⁄9] = 1
Example 9
Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that she gets at least one head?
Solution:
1. Let us denote the sides:
• Head of Re 1 is 1H • Tail of Re 1 is 1T
• Head of Re 2 is 2H •Tail of Re 2 is 2T
2. Let us write the outcomes:
• Outcome 1: Re 1 lands with H on upper face and Re 2 also lands with H on upper face
♦ That is., we have: [1H, 2H]
• Outcome 2: Re 1 lands with T on upper face and Re 2 also lands with T on upper face
♦ That is., we have: [1T, 2T]
• Outcome 3: Re 1 lands with H on upper face and Re 2 lands with T on upper face
♦ That is., we have: [1H, 2T]
• Outcome 4: Re 1 lands with T on upper face and Re 2 lands with H on upper face
♦ That is., we have: [1T, 2H]
• No outcomes other than the above 4 can possibly occur
• We say that: The number of all possible outcomes is 4
• So in the denominator, we will have 4
3. We want to present the probability of 'at least one head'
• We will denote it as P(H)
• If at least one head is obtained, we have an event
4. Towards the first outcome in step(2), we will be writing 'favourable'.
♦ This is because there is at least one head
• Towards the second outcome in step(2), we will be writing 'not favourable'.
♦ This is because there is not even a single head
• Towards the third outcome in step(2), we will be writing 'favourable'.
♦ This is because there is at least one head
• Towards the fourth outcome in step(2), we will be writing 'favourable'.
♦ This is because there is at least one head
5. So we will be having 3 favourable outcomes.
• Thus, in the numerator, we will be having 3
• So P(H) = 3⁄4
■ Can we apply P(H) = [1-P(H)] in this problem?
Let us try:
• (H) indicates an event of 'no head'
• Out of the 4 possible outcomes, only one (outcome 3) is favourable for 'no head'
• So P(H) = 1⁄4
• Thus we get: P(H) = [1-P(H)] = [1 - 1⁄4] = 3⁄4
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