SOLVED PROBABILITY PROBLEMS FOR GRADE 11
Problem 1 :
An integer is chosen at random from the first 100 positive integers. What is the probability that the integer chosen is a prime or multiple of 8?
Solution :
Sample space = {1, 2, 3, 4, ...........100}
n (S) = 100
Let "A" be the event of choosing a number is a prime
A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}
n(A) = 25
P(A) = n(A)/n(S) = 25/100
Let "B" be the event of choosing be a multiple of 8.
B = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96}
n(A) = 12
P(B) = n(B)/n(S) = 12/100
P(A or B) = P(A) + P(B)
= 25/100 + 12/100
= 37 /100
Problem 2 :
A bag contains 7 red and 4 black balls, 3 balls are drawn at random. Find the probability that (i) all are red (ii) one red and 2 black.
Solution :
Total number of balls = 7 red + 4 black
= 11 balls
n(S) = 11C3
= (11⋅10⋅9)/(3⋅2⋅1)
= 165
Let "A" be the event of getting 3 balls are in red color.
n(A) = 7C3
= (7⋅6⋅5)/(3⋅2⋅1)
= 35
P(A) = n(A)/n(S)
= 35/165
P(A) = 7/33
(ii) one red and 2 black.
Let "B" be the event of getting 1 red and 2 black balls.
n(B) = 7C1 ⋅ 4C2
= 7 ⋅ 6
= 42
P(B) = n(B)/n(S)
P(B) = 42/165
P(B) = 14/55
Problem 3 :
A single card is drawn from a pack of 52 cards. What is the probability that (i) the card is an ace or a king (ii) the card will be 6 or smaller (iii) the card is either a queen or 9?
Solution :
Number of cards = 52
n(S) = 52
(i) the card is an ace or a king
Let "A" and "B" be the events of getting an ace and king card respectively.
n(A) = 4 ==> P(A) = n(A) / n(S) = 4/52
n(B) = 4 ==> P(B) = n(B) / n(S) = 4/52
P(A or B) = P(A) + P(B)
= (4/52) + (4/52)
= 8/52
P(A or B) = 2/13
(ii) the card will be 6 or smaller
Let "C" and "D" be the events of getting 6 and smaller than 6 respectively.
Out of 13 spade cards, there will be 1 card numbered 6.
Out of 13 diamond cards, there will be 1 card numbered 6.
Out of 13 clever cards, there will be 1 card numbered 6.
Out of 13 heart cards, there will be 1 card numbered 6.
So, there are 4 cards numbered 6.
n(C) = 4 ==> P(C) = n(C) / n(S) = 4/52
There will be 5 cards in each set of cards.
n(D) = 16 ==> P(D) = n(D) / n(S) = 16/52
P(C or D) = P(C) + P(D)
= (4/52) + (16/52)
= 20/52
P(C or D) = 5/13
(iii) the card is either a queen or 9?
Let "E" and "F" be the events of getting queen and a card numbered 9 respectively.
There will be 4 queen cards in 52 cards.
n(E) = 4 ==> p(E) = n(E)/n(S) = 4/52
n(F) = 4 ==> p(F) = n(F)/n(S) = 4/52
P(C or D) = P(C) + P(D)
= (4/52) + (4/52)
= 8/52
P(E or F) = 2/13
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