PROBLEMS:
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
6)Two diced are tossed the probability that the total score is a prime number?
Sol:Number of total ways n(S) =6 * 6 =36
E =event that the sum is a prime number.
Then E={(1,1)(1,2)(1,4)(1,6)(2,1)(2,3)(2,5)(3,2)(3,4)(4,1)(4,3)(5,2)(5,6)(6,1)(6,5)}
n(E) =15
P(E) =n(E)/n(S)
= 15/36
P(E) = 5/12
7)Two dice are thrown simultaneously .what is the probability of getting two numbers
whose product is even?
Sol : In a simultaneous throw of two dice ,we have n(S) = 6*6
= 36
E =Event of getting two numbers whose product is even
E ={(1,2)(1,4)(1,6)(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)(3,2)(3,4)(3,6)(4,1)
(4,2)(4,3)(4,4)(4,5)(4,6)(5,2)(5,4)(5,6)(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)}
n(E) = 27
P(E) = n(E)/n(S)
= 27 /36
P(E) =3/4
probability of getting two numbers whose product is even is equals to 3/4.
8)In a lottery ,there are 10 prozes and 25 blanks.A lottery is drawn at random.
what is the probability of getting a prize ?
Sol: By drawing lottery at random ,we have n(S) =10C1+25C1
= 10+25
= 35.
E =event of getting a prize.
n(E) =10C1 =10
out of 10 prozes we have to get into one prize .The number of ways 10C1.
n(E) =10
n(S) =35
P(E) =n(E)/n(S)
=10/35
= 2/7
Probability is 2/7.
9)In a class ,30 % of the students offered English,20 % offered Hindi and 10 %offered Both.If a student is offered at random, what is the probability that
he has offered English or Hindi?
Sol: English offered students = 30 %.
Hindi offered students = 20%
Both offered students = 10 %
Then only English offered students E =30 - 10 = 20 %
only Hindi offered students S = 20 -10 % = 10 %
All the students = 100% = E + S + E or S
100 = 20 + 10 + E or S + E and S
Hindi or English offered students =100 - 20 - 10 - 1 = 60 %
Probability that he has offered English or Hindi = 60/100 = 2/5
