Question

a box contains 9 red balls and 6 blue balls. if two are drawn in succession, what is the probability that one of them is red and the other is blue?

Answer 1

Answer:

Probability of getting one red and one blue.

Two cases arises :

Case 1. Red ball in first draw and blue ball in second draw.

So required probability is

$\frac{9}{15} \times \frac{6}{15}$

Case 2. Blue ball in first draw and red ball in second draw.

So required probability is

$\frac{6}{15} \times \frac{9}{15}$

So total probability is

$\frac{9}{15} \times \frac{6}{15} + \frac{6}{15} \times \frac{9}{15} \\ = \frac{12}{25}$

Answer 2

Answer:

1 is the required probability that one of them is red and other is blue .

Step-by-step explanation:

Explanation:

Given, a box contain 9 red balls and 6 blue balls .

Let total number in box is 9+6 =15

( 9 red balls and 6 blue balls )

Step1:

Total number of red balls in box = 9

Total number of blue balls in box = 6

Let A be event of red balls and

let B be the event of blue balls

Probability of red balls  P(A) = $\frac{9}{15}$

Probability of blue balls P(B) = $\frac{6}{15}$

Step2:

Therefore ,

Probability of one red ball and one blue ball

P = P(A)+P(B)

  = $\frac{9}{15}+\frac{6}{15}$ = $\frac{15}{15} = 1$

Final answer :

Hence , the probability that one ball is red and the other is blue is 1 .