2018 VJC P2 Q8

2018 VJC P2 Q8

12 marks
Prelims

The probability of obtaining a ‘\(6\)’ on a biased cubical dice is thrice the probability of rolling every other number on the dice.

Show that the probability of obtaining a ‘\(6\)’ on the biased dice is \(\frac{3}{8}\).[1]

This biased dice is put into a bag together with \(3\) fair cubical dice.

  1. One of the dice is chosen randomly from the bag and rolled once. Find the probability of obtaining a ‘\(6\)’.[2]
  2. One of the dice is chosen randomly and is rolled \(n\) times.
    1. Find the probability that ‘\(6\)’ is obtained on all the \(n\) rolls.[1]
    2. Given that ‘\(6\)’ is obtained for all the \(n\) rolls, the probability that the biased dice is chosen is more than \(0.95\). By forming an inequality in terms of \(n\), solve for the least value of \(n\).[3]
  3. One of the dice is chosen randomly from the bag and rolled once. The dice is then put back into the bag. Another dice is chosen randomly from the bag and rolled once. For every ‘\(6\)’ obtained, the player gains \(\$2\), otherwise the player loses
    \(\$k\). If the game is fair, that is, expected winnings of the player is \(\$0\), show that the value of \(k\) is \(0.56\).[3]
    Find the variance of the player’s winnings.[2]

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Answer:(i) \({0.21875}\) or \({\left(\frac{7}{32}\right)}\) (ii) (a) \({\mathrm{P}\text{ (obtaining}\hspace{0.5em} n\text{ `}\hspace{0.5em}6\text{'s)} = \frac{1}{4} \times \left(\frac{3}{8}\right)^n + \frac{3}{4} \times \left(\frac{1}{6}\right)^n}\) (b) \({5}\) (iii) \({\$^2 2.24}\)

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