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Question
In binomial distribution with five Bernoulli’s trials, the probability of one and two success are 0.4096 and 0.2048 respectively. Find the probability of success.
Solution
Given: X ~ B(n = 5, p)
The probability of X successes is
P(X = x) = `"^nC_x p^x q^(n - x)`, x = 0, 1, 2,...,n
i.e. P(X = x) = `"^5C_x p^x q^(5 - x)`, x = 0, 1, 2, 3, 4, 5
Probabilities of one and two successes are
P(X = 1) = `"^5C_1 p^1 q^(5 - 1)`
and P(X = 2) = `"^5C_2 p^2 q^(5 - 2)` respectively
Given: P(X = 1) = 0.4096 and P(X = 2) = 0.2048
∴ `("P"("X" = 2))/("P"("X" = 1)) = 0.2048/0.4096`
i.e. `(""^5C_2 p^2 q^(5 - 2))/("^5C_1 p^2 q^(5 - 1)) = 1/2`
i.e. `2 xx ""^5C_2 p^2 q^3 = 1 xx ""^5C_1 "pq"^4`
i.e. `2 xx (5 xx 4)/(1 xx 2) xx "p"^2 "q"^3 = 1 xx 5 xx "pq"^4`
i.e. `20 "p"^2"q"^3 = 5"pq"^4`
i.e. 4p = q
i.e. 4p = 1 - p
i.e. 5p = 1
∴ p = `1/5`
Hence, the probability of success is `1/5`.
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Solution:
A pair of dice is thrown 3 times.
∴ n = 3
Let x = number of success (doublets)
p = probability of success (doublets)
∴ p = `square`, q = `square`
∴ x ∼ B (n, p)
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Probability of getting at least two success means x ≥ 2.
∴ P(x ≥ 2) = P(x = 2) + P(x = 3)
= `square` + `square`
= `2/27`
