Exam 7: Discrete Probability

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The probability that the first card has a number on it is 36/52, since 4 · 9 = 36 cards have numbers. At that point the deck has 51 remaining cards, and 3 of them have the same number as the first card drawn. Therefore the final answer is (36/52)(3/51) = 36/884.

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(a) Half the time we select the first urn, in which case the probability that the two balls are both red is $( 6 / 9 ) ( 5 / 8 ) = 5 / 12$ . Half the time we select the second urn, in which case the probability that the two balls are both red is $( 5 / 13 ) ( 4 / 12 ) = 5 / 39$ . Therefore the answer is
$$\frac { 1 } { 2 } \cdot \frac { 5 } { 12 } + \frac { 1 } { 2 } \cdot \frac { 5 } { 39 } = \frac { 85 } { 312 }$$ (b) Let $F$ be the event that the first ball is red, and let $S$ be the event that the second ball is red. We are asked for $p ( S \mid F )$ . By definition, this is $p ( S \cap F ) / p ( F )$ . In part (a) we found that $p ( S \cap F ) = 85 / 312$ . By a simpler calculation, we see that
$$p ( F ) = \frac { 1 } { 2 } \cdot \frac { 6 } { 9 } + \frac { 1 } { 2 } \cdot \frac { 5 } { 8 } = \frac { 31 } { 48 }$$ Thus the answer is
$$\frac { 85 / 312 } { 31 / 48 } = \frac { 170 } { 403 }$$

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The expected value is $\frac { 1 } { 6 } \cdot 1 + \frac { 2 } { 6 } \cdot 2 + \frac { 3 } { 6 } \cdot 3 = \frac { 14 } { 6 }$ . To compute the variance we compute $E \left( X ^ { 2 } \right) - E ( X ) ^ { 2 }$ , where $X$ is the value on the slip. We have $E \left( X ^ { 2 } \right) = \frac { 1 } { 6 } \cdot 1 + \frac { 2 } { 6 } \cdot 4 + \frac { 3 } { 6 } \cdot 9 = \frac { 36 } { 6 } = 6$ , so $V ( X ) = 6 - \left( \frac { 14 } { 6 } \right) ^ { 2 } = \frac { 5 } { 9 }$ .