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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 atleast two-third have newspaper reading habit
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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(at least two third have newspaper reading habit)
`"P"(x ≥ 9 xx 2/3)`
= `"P"(x ≥ 6)`
= P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9)
= `9"c"_6 (2/5)^6 (3/5)^(9 - 6) + 9"c"_7 (2/7)^7 (3/5)^(9 - 7) + 9"c"_8 (2/5)^8 (3/5)^(9 - 8) + 9"c"_9 (2/5)^9 (3/5)^(9 - 9)`
= `9"c"_3 (2/5)^6 (3/5)^3 + 9"c"_2 (2/5)^7 (3/5)^2 + 9"c"_1 (2/5)^8 (3/5)^1 + 1(2/5)^9 (3/5)^0`
= `(9 xx 8 xx 7)/(1 xx 2 xx 3) xx [((2)^6 xx (3)^3)/(5)^9] + (9 xx 8)/(1 xx 2) [((2)^7 xx (3)^2)/(5)^9] + 9 xx [((2)^8 xx 3)/(5)^9] + (2)^9/(5)^9`
= `84 xx ((64 xx 27)/(5)^9) + 36 [(128 xx 9)/(5)^9] + 9 xx [(256 xx 3)/(5)^9] + [512/(5)^9]`
= `145152/(5)^9 + 41472/(5)^9 + 6912/(5)^9 + 512/(5)^9`
= `(145152 + 41472 + 6912 + 512)/(5)^9`
= `194048/195312`
= 0.09935
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