# Find the probability that the visitor obtains answer yes from at least 2 pupils: a. when the number of pupils questioned remains at 4. b. when the number of pupils questioned is increased to 8. - Mathematics and Statistics

Sum

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:
a. when the number of pupils questioned remains at 4.
b. when the number of pupils questioned is increased to 8.

#### 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.

Concept: Binomial Distribution
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 12th Standard HSC Maharashtra State Board
Chapter 8 Binomial Distribution
Miscellaneous exercise 8 | Q 14.2 | Page 255