An-unfair-coin-with-the-probability-of-getting-head-in-one-toss-1-5-If-coin-tosses-n-times-the-probability-of-getting-2-heads-is-equal-to-the-probability-of-getting-3-heads-Find-n-

Question Number 43659 by Joel578 last updated on 13/Sep/18

An unfair coin with the probability of getting

head in one toss = \frac{1}{5} .

If coin tosses n times, the probability of

getting 2 heads is equal to the probability of

getting 3 heads . Find n

Commented by math1967 last updated on 14/Sep/18

n = 14 ?

Commented by Joel578 last updated on 17/Sep/18

Please explain the working Sir

Answered by math1967 last updated on 17/Sep/18

p r o b a b i l i t y g e t t i n g h e a d = \frac{1}{5}

p r o b a b i l i t y o f g e t t i n g t a i l = (1 - \frac{1}{5}) = \frac{4}{5}

f r o m t h e p r o b l e m o f b i n o m i a l d i s t r i b u t i o n

{\overset{n}{c}}_{2} {(\frac{1}{5})}^{2} \times {(\frac{4}{5})}^{n - 2} = \overset{n}{} c_{3} {(\frac{1}{5})}^{3} \times {(\frac{4}{5})}^{n - 3}

\frac{n!}{2! (n - 2)!} \times {(\frac{4}{5})}^{n - 2 - n + 3} = \frac{n!}{3! (n - 3)!} \times {(\frac{1}{5})}^{3 - 2}

\frac{4}{10 (n - 2)} = \frac{1}{30} \Rightarrow 10 n = 120 + 20

∴ n = 14

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