Question

Determine P: A coin is tossed three times, where

1. E: heads on third toss,F:heads on first two tosses.
2. E: at least two heads,F:at most two heads.
3. E: at most two tails,F:at least one tail.

Answer 1

A coin tossed three times, i.e., S = (TTT, HTT, THT, TTH, HHT, HTH, THH, HHH)
⇒ n(S)=8

(i) E: heads on third toss E=(TTH, HTH, THH, HHH)
⇒N(E)=4

$P(E)=\frac {n(E)}{n(S)}$ = $\frac 48$ =$\frac 12$

F: heads on first two toss F=(HHT, HHH)
⇒ n(F)=2

$P(F)=\frac {n(F)}{n(S)}$ =$\frac 28$ =$\frac 14$

∴E∩F = (HHH)
⇒ n(E∩F) = 1

∴$P(E∩F)=\frac {n(E∩F)}{n(S)} =\frac 18$

And, $P(E|F)=\frac {P(E∩F)}{P(F)} =\frac {1/8}{2/8} =\frac 12$

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(ii )E: at least two heads E=(HHT, HTH, THH, HHH)
⇒n(E)=4

$P(E)=\frac {n(E)}{n(S)}$ =$\frac 48$ =$\frac 12$

F: at most two heads F=(TTT, HTT, THT, TTH, HHT, HTH, THH)
⇒n(F)=7

$P(F)=\frac {n(F)}{n(S)}$ =$\frac 78$

∴E∩F = (HHT, HTH, THH)
⇒n(E∩F) = 3

$∴P(E∩F) = \frac {n(E∩F)}{n(S)}$ =$\frac 38$

And, $P(E|F)=\frac {P(E∩F)}{P(F)} =\frac {3/8}{7/8} =\frac 37$

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(iii)E: at most two tails E=(HTT,THT,TTH,HHT,HTH,THH,HHH)
⇒n(E)=7

$P(E)=\frac {n(E)}{n(S)}$ =$\frac 78$

F:at least one tail F = (TTT, HTT, THT, TTH, HHT, HTH, THH)
⇒n(F)=7

$P(F)=\frac {n(F)}{n(S)}$ =$\frac 78$

∴E∩F=(HTT, THT, TTH, HHT, HTH, THH)
⇒n(E∩F)=6

∴$P(E∩F)=\frac {n(E∩F)}{n(S)}$ =$\frac 68$

And, $P(E∩F)=\frac {n(E∩F)}{n(S)}$=$\frac {6/8}{7/8}$ =$\frac 67$