(a) A fair die with six faces is thrown six times. Calculate, correct to three decimal places, the probability of obtaining : (i) exactly three sixes ; (ii)...
Question 1 Report
(a) A fair die with six faces is thrown six times. Calculate, correct to three decimal places, the probability of obtaining :
(i) exactly three sixes ; (ii) at most three sixes.
(b) Eight percent of screws produced by a machine are defective. From a random sample of 10 screws produced by the machine, find the probability that :
(i) exactly two will be defective ; (ii) not more than two will be defective.
Answer Details
Both parts use the binomial distribution \(P(X=r) = \binom{n}{r}p^r(1-p)^{n-r}\).
(a) Die thrown \(n = 6\) times, \(p = \tfrac16\) for a six, \(q = \tfrac56\).
(i) Exactly three sixes:
\[\binom{6}{3}\left(\tfrac16\right)^3\left(\tfrac56\right)^3 = 20\cdot\frac{1}{216}\cdot\frac{125}{216} = \frac{2500}{46656} \approx 0.054.\]
(ii) At most three sixes \(= P(0)+P(1)+P(2)+P(3)\):
\[P(0)=0.334898,\ P(1)=0.401878,\ P(2)=0.200939,\ P(3)=0.053584.\]
Sum \(= 0.991\) (to three decimal places).
(b) Screws: \(n = 10\), \(p = 0.08\) defective, \(q = 0.92\).
(i) Exactly two defective:
\[\binom{10}{2}(0.08)^2(0.92)^8 = 45(0.0064)(0.513219) \approx 0.148.\]
(ii) Not more than two \(= P(0)+P(1)+P(2)\):
\[P(0)=(0.92)^{10}=0.434390,\ P(1)=10(0.08)(0.92)^9=0.377730,\ P(2)=0.147807.\]
Sum \(= 0.960\) (to three decimal places).
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