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Question
In a particular university 40% of the students are having newspaper reading habit. Nine university students are selected to find their views on reading habit. Find the probability that none of those selected have newspaper reading habit
Solution
Let p to the probability of having newspaper reading habit
p = `40/100 = 2/5`
q = 1 – p
= `1 2/5`
= `(5 - 2)/5`
= `3/5` and n = 9
In the binomial distribution p(x = 4) = ncx pxqn-r
The binomial distribution P(x) = `9"C"_x (2/5)^x (3/5)^(9 - x)`
P(none of those selected have newspaper reading
= P(X = 0)
= `9"C"_0 (2/5)^0 (3/5)^(9 - 0)`
= `(1)(1)(3/5)^9`
= `3^9/5^9`
= `(27 xx 27 xx 27)/(125 xx 125 xx 125)`
= `21141/1953125`
= 0.0108
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