**There are 6 red balls,5 blue balls,and 4 green balls in a bag. What is the probability of drawing at least 3 red ball in 5th attempt. Assume balls are drawn with replacement**

I tried to use binomial concept and conditional probability formula. As probability of drawing red ball is 6/17,then took case for if balls drawn are 3red & 4red & 5red & 6red. & Then summed all probability but it got very lengthy and messed up.

Mathematics Asked by Rajakr on December 30, 2020

1 AnswersIf you try to sum up the probabilities of drawing **exactly** $3$ red balls on the third, fourth or fifth attempt, you will find it a bit tricky because you'll always have to keep in mind that the last trial $(say, k),$**must** be a win, and work out the combinations for $k-1$ trials followed by a win on the $k_{th}$ trial, which can become messy.

But if you understand that in $5$ attempts, you can have **at most $2$ failures,** then it becomes very simple.

$dbinom50 q^0p^5 + dbinom51 q^1p^4 +dbinom52 q^2p^3$, and evaluate

Answered by true blue anil on December 30, 2020

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