Question

There are 7 black balls and 8 red balls in an urn. If 4 balls are drawn without replacement, what is the probability that exactly 1 black ball is drawn? Express your answer as a fraction or a decimal number rounded to four decimal places.

Answer 1

Answer:

0.0718

Step-by-step explanation:

Given that there are 7 black balls and 8 red balls.

Probability of an event E is given as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

The required probability that exactly one black ball is drawn out of 4 drawn balls = Probability of drawing 1 black ball $\times$ Probability of drawing 3 black balls.

First of all, let us see the probability of drawing one black ball:

Number of favorable cases = 7 (Total number of black balls)

Total number of cases = 15 (Total number of black balls and red balls)

$P(Black\ Ball) = \dfrac{7}{15}$

Now, have a look at probability of choosing 3 red balls:

We have to choose 3 red balls out of 8 balls:

Total number of favorable cases = $8_{C_3}$ = 8 $\times 7 \times 6$

Now, one black ball is already chosen, so total 14 balls are left out of which 3 balls are to be chosen.

So, total number of cases = ${14}_{C_3} = 14 \times 13 \times 12$

$P(3\ red \ balls) = \dfrac{8 \times 7 \times 6}{14 \times 13 \times 12}$

The required probability =

$P(Black\ balls) \times P(3\ red\ balls)\\\Rightarrow \dfrac{7}{15} \times \dfrac{8 \times 7 \times 6 }{14 \times 13 \times 12}\\\Rightarrow 0.0718$

So, the required probability is 0.0718.