December 2020
Monday 28
28 A bag contains 6 green
and 5 red balls. Another bag B
contain 4 green and 10 red balls
A bag is transferreed from bag
to bag B. Then a ball is drawn
from bag B. Find the probability
that the ball drawn will be red.
29
Tuesday
Answers
Given,
bag A contains 6 green and 5 red balls
bag B contains 4 green and 10 red balls.
Probability = number of favorable cases/total number of cases.
Probability of drawing a green ball = no. of green balls/Total no. of balls in bag
Probability of drawing a red ball = no. of red balls/Total no. of balls in bag
The process is,
first we have to draw a ball from bag A and transfer it to bag B then, again we have to draw a ball from bag B.
And this final drawn ball is red.
Suppose a green ball is transferred from bag A to B .
Probability of drawing a green ball from bag A = 6/11
This ball is transferred to bag B
Now bag B contains 5 green 10 red balls.
Probability of getting a red ball from bag B = 10/15 = 2/3
So, if ball drawn is green in bag A, the probability of getting a red ball is
Probability of drawing a green ball from bag A x Probability of getting a red ball from bag B.
⇒ 6/11 * 2/3 = 4/11
Suppose if a red ball is transferred from bag A to B .
Probability of drawing a red ball from bag A = 5/11
This ball is transferred to bag B
Now bag B contains 4 green and 11 red balls.
Probability of getting a red ball from bag B = 11/15
So, if ball drawn is red in bag A, the probability of getting a red ball is
Probability of drawing a red ball from bag A x Probability of getting a red ball from bag B.
⇒ 5/11 * 11/15 = 1/3
Total probability of getting a red ball in both cases = 4/11 + 1/3
= 23/33
23/33 is the required probability.