Example 1: Two players, Dora and Linda, play a tennis match. It is known that the probability of Dora winning the match is 0.62. What is the probability of Linda winning the match? Solution: Let S and R denote the events that Dora wins the match and Linda wins the match, respectively.
The probability of Dora’s winning =P(S)=0.62 (given)
The probability of Linda’s winning =P(R)=1-P(S)
[As the events R and S are complementary]
=1-0.62=0.38
Ex2: Mandy and Lucy are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year). Solution:
Out of the two friends, one girl, say, Mandy’s birthday can be any day of the year. Now, Lucy’s birthday can also be any day of 365 days in the year.
We assume that these 365 outcomes are equally likely.
(i) If Lucy’s birthday is different from Mandy’s, the number of favorable outcomes for her birthday is 365-1=364
So, P(Lucy’s birthday is different from Mandy’s birthday)=364/365
(ii) P(Mandy and Lucy have the same birthday)
=1-P(both have different birthdays)
=1-364/365=1/365 [Using P(E)=1—P(E)]
Ex3: 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:
There are 40 students, and only one name card has to be chosen.
(i) The number of all possible outcomes is 40
The number of outcomes favorable for a card with the name of a girl =25 (Why?)
Therefore, P(card with name of a girl)=P(Girl)=25/40=⅝
(ii) The number of outcomes favorable for a card with the name of a boy =15 (Why?)
Therefore, P(card with name of a boy)=P(Boy) = 15/40=⅜ Note: We can also determine P(Boy), by taking
P(Boy)=1-P(not Boy)=1-P(Girl)=1-⅝=⅜
Ex4:
Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish (see the given figure).
What is the probability that the fish taken out is a male fish? solution:
Total number of fishes in a tank= Number of male fishes+Number of female fishes =5+8=13
Probability of getting a male fish = 5/13
Ex5. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman? Solution:
There are four men and six women on the city council.
As one council member is to be selected for a committee at random, the sample space contains 10=(4+6) elements.
Let A be the event in which the selected council member is a woman. Accordingly, n(A)=6
Ex6. Experiment:
Pakistan and India play a cricket match. Find the probability in each case: Events Happening:
(i). Pakistan wins Solution:
Since there are three possible, results of the match, Pakistan wins, lose or match tie. So,
Let event A denote that Pakistan will win the match. Then total possible outcomes n(S)=3, favorable outcome n(A)=1.
(ii). India does not lose Solution:
Let event B denote that India will win the match or match will be a tie. Then Total possible outcomes n(S)=3
Favorable outcome n(B)=2
Ex7. Experiment:
One chit out of 30 containing the names of 30 students of a class of 18 boys and 12 girls is taken out at random, for nomination as the monitor of the class. Find the probability in each case:
Events Happening:
(i). The monitor is the boy Solution:
Boys =18; Girls =12. Total =30.
A student is taken as a random. So, possible outcomes
n(S)=30C1=30
Let event A denote that the monitor is boy
(ii). the monitor is a girl Solution:
Let event B denote that the monitor is a girl.
