**Conditional Probability Word Problems :**

Here we are going to see some example problems on conditional probability.

**Question 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) = ^{5}C_{1}/^{8}C_{1} = 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) = ^{4}C_{1}/^{10}C_{1 } = 4/10 ----(2)

(1) ⋅ (2)

P(W/B1) ⋅ P(W/B2) = (5/8) ⋅ (4/10)

= 1/4

(ii) both are black

P(B/B1) = ^{3}C_{1}/^{8}C_{1} = 3/8 ----(3)

P(B/B2) = ^{6}C_{1}/^{10}C_{1 } = 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

**Question 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

**Question 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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