# CONDITIONAL PROBABILITY WORD PROBLEMS

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)  =  4C1/10C =  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)  =  6C1/10C =  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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