Question

A bag contains 4 white and 5 blue balls. Another bag contains 5 white and 7 blue balls. What is the probability of choosing two balls such that one is white and the other is blue

Answer 1

Answer:

$\frac{53}{108}$ is the required probability of one is white and the other is blue

Step-by-step explanation:

Explanation:

Given, A bag contains 4 white and 5 blue balls . Another bag contains 5 white and 7 blue .

Therefore , total number of balls in bag $I$ = 9  

                                                              (4 white balls and 5 blue balls)

and total number of balls in bag $II$ = 12

                                                                  (5 white and 7 blue balls)

Step1:

Let 'A' be the event that balls selected from the first bag is white and

balls selected from the second bag is blue .

and let 'B' be the event that balls selected from the first bag is blue and

balls selected from the second bag is white.

P(A) = $\frac{4}{9}$ ×$\frac{7}{12}$ = $\frac{28}{108}$

(Where $\frac{4}{9}$ is the probability of white balls for first bag and $\frac{7}{12}$is the probability of blue balls of second bag  )

P(B) = $\frac{5}{9}$ ×$\frac{5}{12}$ = $\frac{25}{108}$

(Where $\frac{5}{9}$ is the probability of blue balls for first bag and $\frac{5}{12}$ is the probability of white balls for second bag )

Step 2:

Therefore , probability of choosing two balls such that one is white and other is blue

P = P(A) + P(B)

  = $\frac{28}{108} +\frac{25}{108}$ = $\frac{53}{108}$

Final answer :

Hence, $\frac{53}{108}$  the probability of choosing two balls such that  one is white and the other is blue .