Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to :
Solution
${}^5{C_1}{p^1}{q^4}$ = 0.4096 ..... (1)<br><br>${}^5{C_2}{p^2}{q^3}$ = 0.2048 ..... (2)<br><br>${{(1)} \over {(2)}} \Rightarrow {q \over {2p}} = 2 \Rightarrow q = 4p$<br><br>$p + q = 1 \Rightarrow P = {1 \over 5},q = {4 \over 5}$<br><br>P (exactly 3) = $${}^5{C_3}{(p)^3}{(q)^2} = {}^5{C_3}{\left( {{1 \over 5}} \right)^3}{\left( {{4 \over 5}} \right)^2}$$<br><br>$= 10 \times {1 \over {125}} \times {{16} \over {25}} = {{32} \over {625}}$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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