1)An biased die is tossed.Find the probability of getting a multiple of 3?
Sol: Here we have sample space S={1,2,3,4,5,6}.
Let E be the event of getting a multiple of 3.
Then E={3,6}.
P(E) =n(E)/n(S).
n(E) =2,
n(S) =6.
P(E) =2/6
P(E) =1/3.
2)In a simultaneous throw of a pair of dice,find the probability of getting a total more than 7?
Sol: Here we have sample space n(S) =6*6 =36.
Let E be the event of getting a total more than 7.
={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),
(5,3),(6,2),(4,5),(5,4),
(5,5),(4,6),(6,4)}
n(E) =15
P(E) = n(E)/n(S)
= 15/36.
P(E) = 5/12.
3)A bag contains 6 white and 4 black balls .Two balls are drawn at random .Find the probability that they are of the
same colour?
Sol: Let S be the sample space.
Number of ways for drawing two balls out of 6 white and
4 red balls = 10C2
=10!/(8!*2!)
= 45.
n(S) =45.
Let E =event of getting both balls of the same colour.
Then
n(E) =number of ways of drawing ( 2balls out of 6) or
(2 balls out of 4).
= 6C2 +4C2
= 6!/(4!*2!) + 4!/(2! *2!)
= 6*5/2 +4 *3/2
=15+6 =21.
P(E) =n(E)/n(S) =21/45 =7/45.
4)Two dice are thrown together .What is the probability that the sum of the number on the two faces is divisible by 4 or 6?
Sol: n(S) = 6*6 =36.
E be the event for getting the sum of the number on the two
faces is divisible by 4 or 6.
E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)
(5,1)(5,3)(6,2)(6,6)}
n(E) =14.
Hence P(E) =n(E)/n(S)
= 14/36.
P(E) = 7/18
5)Two cards are drawn at random from a pack of 52 cards What is the probability that either both are black or both are queens?
Sol: total number of ways for choosing 2 cards from
52 cards is =52C2 =52 !/(50!*2!)
= 1326.
Let A= event of getting bothe black cards.
Let B= event of getting bothe queens
AnB=Event of getting queens of black cards
n(A) =26C2.
We have 26 black cards from that we have to choose 2 cards.
n(A) =26C2=26!/(24!*2!)
= 26*25/2=325
from 52 cards we have 4 queens.
n(B) = 4C2
= 4!/(2!* 2!) =6
n(AnB) =2C2. =1
P(A) = n(A) /n(S) =325/1326
P(B) = n(B)/n(S) = 6/1326
P(A n B) = n(A n B)/n(S) = 1/1326
P(A u B) = P(A) +P(B) -P(AnB)
= 325/1326 + 6/1326 -1/1326
= 330/1326
P(AuB) = 55/221
Sol: Here we have sample space S={1,2,3,4,5,6}.
Let E be the event of getting a multiple of 3.
Then E={3,6}.
P(E) =n(E)/n(S).
n(E) =2,
n(S) =6.
P(E) =2/6
P(E) =1/3.
2)In a simultaneous throw of a pair of dice,find the probability of getting a total more than 7?
Sol: Here we have sample space n(S) =6*6 =36.
Let E be the event of getting a total more than 7.
={(1,6),(2,5),(3,4),(4,3)(5,2),(6,1)(2,6),(3,5),(4,4),
(5,3),(6,2),(4,5),(5,4),
(5,5),(4,6),(6,4)}
n(E) =15
P(E) = n(E)/n(S)
= 15/36.
P(E) = 5/12.
3)A bag contains 6 white and 4 black balls .Two balls are drawn at random .Find the probability that they are of the
same colour?
Sol: Let S be the sample space.
Number of ways for drawing two balls out of 6 white and
4 red balls = 10C2
=10!/(8!*2!)
= 45.
n(S) =45.
Let E =event of getting both balls of the same colour.
Then
n(E) =number of ways of drawing ( 2balls out of 6) or
(2 balls out of 4).
= 6C2 +4C2
= 6!/(4!*2!) + 4!/(2! *2!)
= 6*5/2 +4 *3/2
=15+6 =21.
P(E) =n(E)/n(S) =21/45 =7/45.
4)Two dice are thrown together .What is the probability that the sum of the number on the two faces is divisible by 4 or 6?
Sol: n(S) = 6*6 =36.
E be the event for getting the sum of the number on the two
faces is divisible by 4 or 6.
E={(1,3)(1,5)(2,4?)(2,2)(3,5)(3,3)(2,6)(3,1)(4,2)(4,4)
(5,1)(5,3)(6,2)(6,6)}
n(E) =14.
Hence P(E) =n(E)/n(S)
= 14/36.
P(E) = 7/18
5)Two cards are drawn at random from a pack of 52 cards What is the probability that either both are black or both are queens?
Sol: total number of ways for choosing 2 cards from
52 cards is =52C2 =52 !/(50!*2!)
= 1326.
Let A= event of getting bothe black cards.
Let B= event of getting bothe queens
AnB=Event of getting queens of black cards
n(A) =26C2.
We have 26 black cards from that we have to choose 2 cards.
n(A) =26C2=26!/(24!*2!)
= 26*25/2=325
from 52 cards we have 4 queens.
n(B) = 4C2
= 4!/(2!* 2!) =6
n(AnB) =2C2. =1
P(A) = n(A) /n(S) =325/1326
P(B) = n(B)/n(S) = 6/1326
P(A n B) = n(A n B)/n(S) = 1/1326
P(A u B) = P(A) +P(B) -P(AnB)
= 325/1326 + 6/1326 -1/1326
= 330/1326
P(AuB) = 55/221
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