87. In a game a man wins a rupee for a six and looses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amounts he wins/looses.

<p><strong>87. </strong>When a die is thrown, then probability of getting a six = 16</p><p>then, probability of not getting a six = 1 - 16 = 56</p><p>If the man gets a six in the first throw, then</p><p>probability of getting a six = 16</p><p>If he does not get a six in first throw, but gets a six in second throw, then</p><p>probability of getting a six in the second throw = 56×16 = 536</p><p>If he does not get a six in the first two throws, but gets in the third throw, then</p><p>probability of getting a six in the third throw = 56×56×16 = 25216</p><p>probability that he does not get a six in any of the three throws = 56×56×56 = 125216</p><p>In the first throw he gets a six, then he will receive Re 1.</p><p>If he gets a six in the second throw, then he will receive Re (1 - 1) = 0</p><p>If he gets a six in the third throw, then he will receive Rs (-1 - 1 + 1) = Rs (-1), </p><p>that means he will lose Re 1 in this case.</p><p>Expected value = 16×1 + 56×16 × 0 + 56×56×16×-1 = 11216</p><p>So, he will loose Rs 11216.</p>

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