(a) Eight coins are tossed at once. Find, correct to three decimal places, the probability of obtaining : (i) exactly 8 heads ; (ii) at least 5 heads ; (iii...
Question 1 Report
(a) Eight coins are tossed at once. Find, correct to three decimal places, the probability of obtaining :
(i) exactly 8 heads ; (ii) at least 5 heads ; (iii) at most 1 head.
(b) In how many ways can four letters from the word SHEEP be arranged (i) without any restriction ; (ii) with only one E.
Answer Details
(a) Eight coins, \(n = 8\), \(p = 0.5\) for a head, so each outcome count is \(\binom{8}{r}/256\).
(i) Exactly 8 heads: \(\left(\tfrac12\right)^8 = \dfrac{1}{256} \approx 0.004\).
(ii) At least 5 heads \(= \dfrac{\binom{8}{5}+\binom{8}{6}+\binom{8}{7}+\binom{8}{8}}{256} = \dfrac{56+28+8+1}{256} = \dfrac{93}{256} \approx 0.363\).
(iii) At most 1 head \(= \dfrac{\binom{8}{0}+\binom{8}{1}}{256} = \dfrac{1+8}{256} = \dfrac{9}{256} \approx 0.035\).
(b) The word SHEEP has letters S, H, E, E, P (the E repeats). Choose and arrange 4 letters.
(i) Without restriction, split by how many E's are used:
- Both E's plus 2 of \(\{S,H,P\}\): \(\binom{3}{2}\times\dfrac{4!}{2!} = 3\times12 = 36\).
- At most one E, i.e. the four distinct letters \(S,H,E,P\): \(4! = 24\).
Total \(= 36 + 24 = 60\) arrangements.
(ii) With only one E, the other three must be \(S, H, P\), giving four distinct letters: \(4! = 24\) arrangements.
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