A bag contains 24 mangoes out of which six are bad. If 6 mangoes are selected randomly from the bag with replacement, find the probability that not more than 3 are bad.
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Correct Answer: Option
Explanation:
Let. X = Prob. that a good mango is selected = 18/24= 3/4
Y Prob. that a bad mango is selected = 6/24 1/4
Using the binomial probability distribution, we have:
(X+Y)\(^6\) = X\(^6\) + \(^6{C}_1 {X}^5 Y\) + \(^6{C}_2 {X}^4 {Y}^2\) + \(^6{C}_3 {X}^3 {Y}^3\) + \(^6{C}_4 {X}^2 {Y}^4\) + \(^6{C}_5 {X} {Y}^5\) + \(^6{C}_6 {Y}^6\)
Probability that not more than 3 are bad is
= \(^6{C}_1 {X}^5 Y\) + \(^6{C}_2 {X}^4 {Y}^2\) + \(^6{C}_3 {X}^3 {Y}^3\)
= 6(3/4)\(^5\) (1/4) + 15(3/4)\(^4\) (1/4)\(^2\) + 20(3/4)\(^3\) (1/4)\(^3\)
= 6(243/1024)(1/4) + 15(81/256)(1/16) + 20(27/64)(1/64)
= 0.36 + 0.30 + 0.13
≈ 0.79
Let. X = Prob. that a good mango is selected = 18/24= 3/4
Y Prob. that a bad mango is selected = 6/24 1/4
Using the binomial probability distribution, we have:
(X+Y)\(^6\) = X\(^6\) + \(^6{C}_1 {X}^5 Y\) + \(^6{C}_2 {X}^4 {Y}^2\) + \(^6{C}_3 {X}^3 {Y}^3\) + \(^6{C}_4 {X}^2 {Y}^4\) + \(^6{C}_5 {X} {Y}^5\) + \(^6{C}_6 {Y}^6\)
Probability that not more than 3 are bad is
= \(^6{C}_1 {X}^5 Y\) + \(^6{C}_2 {X}^4 {Y}^2\) + \(^6{C}_3 {X}^3 {Y}^3\)
= 6(3/4)\(^5\) (1/4) + 15(3/4)\(^4\) (1/4)\(^2\) + 20(3/4)\(^3\) (1/4)\(^3\)
= 6(243/1024)(1/4) + 15(81/256)(1/16) + 20(27/64)(1/64)
= 0.36 + 0.30 + 0.13
≈ 0.79