Question

A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. If from the selected bag one ball is drawn, then the probability that the ball drawn is red, is

(please explain in detail)​

Answer 1

Given:
In the first bag:
•Number of red balls = 4
•Number of black balls = 3
In the second bag:
•Number of red balls = 2
•Number of black balls = 4

To find: The probability of the ball being drawn is red if a single ball is drawn.

Solution:
•Total number of balls in the first bag: 4+3=7
•Total number of balls in the second bag: 2+4=6
•Probalility of either of the bags being chosen= 1/2
•Probability of choosing a red ball from the
first bag out of 7 balls = 4/7
•Probability of choosing a red ball from the second bag out of 6 balls= 2/6 = 1/3

Therefore, the probability of the ball drawn being red is = total of individual probability x probability of choosing single bag

=1/2[4/7+1/3]
=1/2[12+7/21]
=19/42

Answer= 19/42

Answer 2

7 balls in bag 1, 4 of which are red (4/7)

5 balls in bag 2, 2 of which are red (2/5)

Add these (rather than multiply) because one doesn’t depend on the other. So (4/7)+((2/5)=34/35.

Multiply that result by 1/2 because there’s a 50% chance of which bag is chosen.

Ans. 17/35.