Ex1. From a box containing orange-flavored candies, Billy takes-out one candy without looking. find the probability in each case:

Events happening:

(i). The candy is orange flavored

Solution:

Let event A be the no of occurring that candy is orange flavored.

Total possible outcomes =*n*(S)=1.

Since the box contains just orange flavor candies, so at the possible outcomes are favourable, that is favourable outcomes =*n*(A)=1. So, probability

*P*(A)=

*n*(A)/

*n*(S) =1/1=1

(ii). The candy is lemon flavored

Solution:

Let event B be the no of occurring that candy is lemon flavored.

Total possible outcomes =

*n*(S)=1

Since the box contains just orange flavor candies, so at the possible outcomes are favourable, that is favourable outcomes =

*n*(B)=0. So, probability

*P*(B)=

*n*(B)/

*n*(S) =0/1=0

Ex2. The blood groups of 200 people is distributed as follows: 50 have type A blood, 65 have B blood type, 70 have O blood type and 15 have type AB blood. If a person from this group is selected at random, what is the probability that this person has O blood type?

Solution:

Ex3. In a bundle of 50 shirts, 44 are good, 4 have minor defects and 2 have major defects. What is the probability that:

(i) it is acceptable to a trader who accepts only a good shirt?

(ii) it is acceptable to a trader who rejects only a shirt with major defects?

Solutions:

Total number of shirts =50

Total number of elementary events =50→*n*(S)=50

(i) Since trader accepts only good shirts and number of good shirts =44

Event of accepting good shirts *n*(E)=44

Probability of accepting a good shirt =*n*(E)/*n*(S) =44/50=0.88

(ii) Since trader rejects shirts with major defects only and number of shirts with major defects =2

Event of accepting shirts =50-2=48=*n*(E)

Probability of accepting shirts =*n*(E)/*n*(S) =48/50=0.96

Ex4. A bag contains twenty Rs 5 coins, fifty Rs 2 coins and thirty Re 1 coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin:

(i) will be a Re 1 coin?

(ii) will not be a Rs 2 coin?

(iii) will neither be a Rs 5 coin nor be a Re 1 coin?

Solutions:

Total number of coins =20+50+30=100

Total possible outcomes =100=*n*(S)

(i) Number of favorable outcomes for Re 1 coins =30=*n*(E)

Probability(Re 1 coin) =*n*(E)/*n*(S) =30/100=0.30

(ii) Number of favorable outcomes for not a Rs 2 coins =number of favorable outcomes for Re 1 or Rs 5 coins 30+20=100=*n*(E)

Probability(not Rs 2 coin) =*n*(E)/*n*(S) =50/100=0.50

(iii) number of favorable outcomes for neither Re 1 nor Rs 5 coins =Number of favorable outcomes for Rs 2 coins =50

Probability(neither Re 1 nor Rs 5 coin) =*n*(E)/*n*(S) =50/100=0.50

Ex5. A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be

(i) a blue card (ii) not a yellow card

(iii) neither yellow nor a blue card.

Solutions:

Total number of possible outcomes *n*(S)=100+200+50=350 (100 red, 200 yellow & 50 blue)

(i) E→ event of getting blue card.

Number of favorable outcomes *n*(E)=50 {50 blue cards}

(ii) E→ event of getting a yellow card,

*n*(E)=200 (200 yellow)

Ē→ event of getting a non-yellow card

(iii) getting neither yellow nor a blue card means getting a red card.

*P*(a red)=100/350=2/7