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Question
In a large school, 80% of the pupil like Mathematics. A visitor to the school asks each of 4 pupils, chosen at random, whether they like Mathematics.
Find the probability that the visitor obtains answer yes from at least 2 pupils:
- when the number of pupils questioned remains at 4.
- when the number of pupils questioned is increased to 8.
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Solution
Let X = number of pupils like Mathematics.
p = probability that pupils like Mathematics
∴ p = 80% = `80/100 = 4/5`
and q = 1 – p = `1 - 4/5 = 1/5`
Given: n = 4
∴ X ~ B `(4, 4/5)`
The p.m.f. of X is given by P(X = x) = `"^nC_x p^x q^(n - x)`
i.e. p(x) = `"^4C_x (4/5)^x (1/5)^(4 - x)` x = 0, 1, 2, 3, 4
a. P(visitor obtains the answer yes from at least 2 pupils when the number of pupils questioned remains at 4) = P(X ≥ 2)
= P(X = 2) + P(X = 3) + P(X = 4)
`= ""^4C_2 (4/5)^2 (1/5)^(4 - 2) + ""^4C_3 (4/5)^3 (1/5)^(4 - 3) + "^4C_4 (4/5)^4 (1/5)^(4 - 4)`
`= (4 xx 3)/(1 xx 2) xx 16/5^2 xx 1/5^2 + 4 xx 64/5^3 xx 1/5 + 1 xx 256/5^4`
`= 96/5^4 + 256/5^4 + 256/5^4`
`= (96 + 256 + 256)1/5^4`
`= 608/5^4 = 608/625`
b. P(the visitor obtains the answer yes from at least 2 pupils when number of pupils questioned is increased to 8)
= P(X ≥ 2)
= 1 – P(X < 2)
= 1 – [P(X = 0) + P(X = 1)]
= `1 - [""^8C_0 (4/5)^0 (1/5)^(8 - 0) + ""^8C_1 (4/5)^1 (1/5)^(8 - 1)]`
= `1 - [1 (1) (1/5)^8 + (8)(4/5)(1/5)^7]`
= `1 - [1/5^8 + 32/5^8]`
= `1 - 33/5^8`.
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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)
P(x) = nCxpx qn–x
Probability of getting at least two success means x ≥ 2.
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= `square` + `square`
= `2/27`
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