(a) In a school, the ratio of those who passed to those who failed in a History test is 4 : 1. If 7 students are selected at random from the school, find, c...
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(a) In a school, the ratio of those who passed to those who failed in a History test is 4 : 1. If 7 students are selected at random from the school, find, correct to two decimal places, the probability that :
(i) at least 3 passed the test ; (ii) between 3 and 6 students failed the test.
(b) A fair die is thrown five times; find the probability of obtaining a six three times.
Answer Details
(a) Pass:fail \(=4:1\), so \(P(\text{pass})=0.8,\ P(\text{fail})=0.2\), with \(n=7\).
(i) At least 3 passed. Let \(X=\) number who pass, \(p=0.8\).
\[P(X\ge3)=1-P(0)-P(1)-P(2)\]
\(P(0)=(0.2)^{7}=0.0000128\), \(P(1)=7(0.8)(0.2)^{6}=0.000358\), \(P(2)=21(0.8)^{2}(0.2)^{5}=0.004301\).
\[P(X\ge3)=1-0.004672=0.995\approx1.00\]
(ii) Between 3 and 6 failed (i.e. \(4\) or \(5\) failed). Let \(Y=\) number who fail, \(p=0.2\).
\(P(Y=4)=\binom{7}{4}(0.2)^{4}(0.8)^{3}=35(0.0016)(0.512)=0.028672\).
\(P(Y=5)=\binom{7}{5}(0.2)^{5}(0.8)^{2}=21(0.00032)(0.64)=0.004301\).
\[P(4\le Y\le5)=0.028672+0.004301=0.032973\approx0.03\]
(b) A fair die thrown 5 times, \(P(\text{six})=\tfrac{1}{6}\). Probability of exactly three sixes:
\[\binom{5}{3}\left(\tfrac{1}{6}\right)^{3}\left(\tfrac{5}{6}\right)^{2}=10\cdot\frac{1}{216}\cdot\frac{25}{36}=\frac{250}{7776}\approx0.03\]
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