An event that is affected by the occurrence of previous events is known as a dependent event. For example, if a colored ball has to be taken out from a bag having balls of different colors and no replacement is done then each time the ball is taken out is an event dependent on the outcome of previous events. For two events A and B if A has occured before B, then the probability of the dependent event $B$ = $P(\frac{B}{A})$.

## Word Problems

**Problem 1:**

If there are 10 girls and 8 boys in a class, and two students will be chosen one after another for debate competition then find the probability that both are boys.

Solution:

Probability of boy1 and boy2 = $P(boy1) \times P$ $(\frac{boy2}{boy1})$

P(boy1 and boy2) = $\frac{8}{18}$ $\times$ $\frac{7}{17}$ = $\frac{28}{153}$

When first boy is chosen, probability will be for choosing from 8 favorable events from a sample space of 18.

When the second boy is chosen, favorable events become 7 and sample space is of 17.**Problem 2:**

2 out of 5 bulbs bought in a house are defective. if two bulbs are tested, then find the probability that both are defective.

**Solution: **

Probability of first bulb being defective = $\frac{2}{5}$

Probability of second bulb being defective when first has already been defective = $\frac{1}{4}$

Probability of both the bulb being defective = $\frac{2}{5}$ $\times$ $\frac{1}{4}$ = $\frac{1}{10}$**Problem 3:**

Two cards are chosen from a deck of 52 cards without replacement. Find the probability that one is red and second is black.

**Solution: **

Probability of first card being red = $\frac{26}{52}$ = $\frac{1}{2}$

Probability of second card being black when first is chosen red = $\frac{26}{51}$

Probability of one card being red and other being black = $\frac{1}{2}$ $\times$ $\frac{26}{51}$ = $\frac{26}{102}$**Problem 4:**

From a deck of 52 cards what is the probability of choosing two kings?

Solution:

Probability of getting first king = $\frac{4}{52}$ = $\frac{1}{13}$

Probability of getting second king = $\frac{3}{51}$

Probability of getting both kings = $\frac{3}{13\times 51}$ = $\frac{1}{221}$**Problem 5:**

3 out of 10 kids use car to come to school in Delhi. If two kids are surveyed, what is the probability that both do not use car?

Solution:

Probability of first kid not coming by car = $\frac{7}{10}$

Probability of second kid not coming by car = $\frac{6}{9}$ = $\frac{2}{3}$

Probability of both kids not coming by car = $\frac{7}{10}$ $\times$ $\frac{2}{3}$ = $\frac{7}{15}$