API- Ace Petroleum Engineering
There r 3 bags containing 2 green+4 blue ball; 5 green+ 4 blue balls; 3 green+6 blue balls. Find the probability that the ball chosen randomly from any of the bags is a green ball ?
#MathsSolution
Correct answer is GIVEN,
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There will be 3 cases.
Case 1 : Ball is chosen from 1st bag.
Probability of green ball from 1st bag = (1/3)×(2/6)
= 1/9
Case 2 : Ball is chosen from 2nd bag.
Probability of green ball from 2nd bag = (1/3)×(5/9)
= 5/27
Case 3 : Ball is chosen from 3rd bag.
Probability of green ball from 3rd bag
= (1/3)×(3/9)
= 1/9
So required probability
= 1/9 + 5/27 + 1/9
= 11/27
Correct answer is GIVEN,
.
.
.
.
.
.
.
.
.
.
.
There will be 3 cases.
Case 1 : Ball is chosen from 1st bag.
Probability of green ball from 1st bag = (1/3)×(2/6)
= 1/9
Case 2 : Ball is chosen from 2nd bag.
Probability of green ball from 2nd bag = (1/3)×(5/9)
= 5/27
Case 3 : Ball is chosen from 3rd bag.
Probability of green ball from 3rd bag
= (1/3)×(3/9)
= 1/9
So required probability
= 1/9 + 5/27 + 1/9
= 11/27