\({A}\), \({B}\) and \({C}\) are taking part in a game in which a prize is hidden in one of \(n\) identical closed boxes, where \(n > 4\). First, \({A}\) opens at random two boxes in an attempt to find the prize. If \({A}\) fails to find the prize, \({B}\) then opens at random two of the remaining boxes in an attempt to find the prize. If \({B}\) fails to find the prize, \({C}\) opens all the remaining boxes and finds the prize.
Find, in terms of \(n\), the probability that \({A}\) finds the prize.[1]
Show that the probability that B finds the prize is \(\frac{2}{n}\).[1]
Find the range of values of \(n\) for which the probability that \({C}\) finds the prize is more than \(8\) times the probability that either \({A}\) or \({B}\) find the prize.[3]
The events \(Q\) and \(R\) are such that \({P}\left( Q\cap R \right)=x\), \(\mathrm{P}\left( Q\cap {R}' \right)=0.1\), \({P}\left( {Q}'\cap R \right)=\mathrm{P}\left( Q \right)\) and \({P}\left( \left( Q\cup R \right)' \right)=\mathrm{P}\left( Q \right)+2\mathrm{P}\left( R \right)\). By using a Venn diagram, or otherwise, find the exact value of \(x\).[3]
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Default solution
Answer:(a)(i) \(\frac2n\) (ii) \(\frac2n\) (iii) \(n>36\) (b) \(x=\frac1{14}\)