# Probability of getting one of various items

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