Problem 1 :
One bag contains 5 white and 3 black balls. Another bag contains 4 white and 6 black balls. If one ball is drawn from each bag, find the probability that (i) both are white (ii) both are black (iii) one white and one black
Solution :
Let "w" be the event of choosing white ball.
Let "B1" and "B2" be the events of choosing first and second bag respectively.
Bag 1 :
There are 5 white and 3 black balls.
Number of ball in bag 1 = 8
Probability of selecting white ball from bag 1
P(W/B1) = 5C1/8C1 = 5/8 ----(1)
Bag 2 :
There are 4 white and 6 black balls.
Number of ball in bag 2 = 10
Probability of selecting white ball from bag 2
P(W/B2) = 4C1/10C1 = 4/10 ----(2)
(1) ⋅ (2)
P(W/B1) ⋅ P(W/B2) = (5/8) ⋅ (4/10)
= 1/4
(ii) both are black
P(B/B1) = 3C1/8C1 = 3/8 ----(3)
P(B/B2) = 6C1/10C1 = 6/10 ----(4)
(3) ⋅ (4)
P(B/B1) ⋅ P(B/B2) = (3/8) ⋅ (6/10)
= 9/40
(iii) one white and one black
= P(W/B1) ⋅ P(B/B2) + P(B/B1) ⋅ P(W/B2)
= (5/8)(6/10) + (3/8)(4/10)
P(one white and one black) = 21/40
Problem 2 :
Two thirds of students in a class are boys and rest girls. It is known that the probability of a girl getting a first grade is 0.85 and that of boys is 0.70. Find the probability that a student chosen at random will get first grade marks
Solution :
P(boys) = 2/3
P(Girls) = 1/3
Let "A" be the event of getting first grade
P(Boy getting first grade) = P(A/B) = 0.70 and
P(Girl getting first grade) = P(A/G) = 0.85
P(Selecting a student getting first grade)
= P(B) P(A/B) + P(G) P(A/G)
= (2/3) (0.70) + (1/3) (0.85)
= (140/300) + (85/300)
= 225/300
= 3/4
= 0.75
Problem 3 :
Given P(A) = 0.4 and P(AUB) = 0.7. Find P(B) if (i) A and B are mutually exclusive (ii) A and B are independent events (iii) P(A / B) = 0.4 (iv) P(B / A) = 0.5
Solution :
(i) A and B are mutually exclusive
If A and B are mutually exclusive events, then P(AnB) = 0
P(AUB) = P(A) + P(B) - P(AnB)
0.7 = 0.4 + P(B)
P(B) = 0.7 - 0.4 = 0.3
(ii) A and B are independent events
P(AnB) = P(A) P(B)
P(AUB) = P(A) + P(B) - P(A) P(B)
0.7 = 0.4 + P(B) (1 - P(A))
0.7 - 0.4 = P(B) (1 - 0.4)
0.3 = P(B) (0.6)
P(B) = 0.3/0.6
= 0.5
(iii) P(A / B) = 0.4
P(A/B) = P(AnB)/P(B)
0.4 = P(AnB)/P(B)
P(AnB) = 0.4 P(B) ----(1)
P(AUB) = P(A) + P(B) - P(AnB)
0.7 = 0.4 + P(B) - 0.4 P(B)
0.3 = P(B)(1- 0.4)
0.3 = P(B)(0.6)
P(B) = 0.3/0.6 = 0.5
(iv) P(B/A) = 0.5
P(B/A) = P(AnB)/P(A)
0.5 = P(AnB)/0.4
P(AnB) = 0.2
P(AUB) = P(A) + P(B) - P(AnB)
0.7 = 0.4 + P(B) - 0.2
0.7 = 0.2 + P(B)
0.7 - 0.2 = P(B)
P(B) = 0.5
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