Probability Sample Questions And Answers

Probability Sample Questions And Answers

• Posted by  Geetha
29 Jan, 2012

Ex 1. In a throw of a coin ,find the probability of getting a head.
sol. Here s={H,T} and E={H}.
P(E)=n(E)/n(S)=1/2

Ex2.Two unbiased coin are tossed .what is the probability of getting atmost one head?
sol.Here S={HH,HT,TH,TT}
Let Ee=event of getting one head
E={TT,HT,TH}
P(E)=n(E)/n(S)=3/4

Ex3.An unbiased die is tossed .find the probability of getting a multiple of 3
sol. Here S={1,2,3,4,5,6}
Let E be the event of getting the multiple of 3
then ,E={3,6}
P(E)=n(E)/n(S)=2/6=1/3

Ex4. In a simultaneous throw of pair of dice .find the probability of getting the total more than 7
sol. Here n(S)=(6*6)=36
let E=event of getting a total more than 7
={(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}
P(E)=n(E)/n(S)=15/36=5/12.

Ex5. A bag contains 6 white and 4 black balls .2 balls are drawn at random. find the probability that they are of same colour.
Sol .let S be the sample space
Then n(S)=no of ways of drawing 2 balls out of (6+4)=10c2=(10*9)/(2*1)=45
Let E=event of getting both balls of same colour
Then n(E)=no of ways(2 balls out of six) or(2 balls out of 4)
=(6c2+4c2)=(6*5)/(2*1)+(4*3)/(2*1)=15+6=21
P(E)=n(E)/n(S)=21/45=7/15

Ex6.Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6
sol. Clearly n(S)=6*6=36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.Then

E={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),
(6,6)}
n(E)=14.
Hence p(e)=n(e)/n(s)=14/36=7/18

Ex7.Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?
sol. We have n(s)=52c2=(52*51)/(2*1)=1326.
Let A=event of getting both black cards
B=event of getting both queens
A∩B=event of getting queen of black cards
n(A)=26c2=(26*25)/(2*1)=325,
n(B)=4c2=(4*3)/(2*1)=6 and
n(A∩B)=2c2=1
P(A)=n(A)/n(S)=325/1326;
P(B)=n(B)/n(S)=6/1326 and
P(A∩B)=n(A∩B)/n(S)=1/1326
P(A∪B)=P(A)+P(B)-P(A∩B)=(325+6-1/1326)=330/1326=55/221