![]() ![]() Here, $E=2+EX$ because, if the first two tosses are $HT$, we have wasted two coin tosses and we start over at the third toss.īy letting $E=2+(1-p)\mu$ in Equation 5. We still need to find $E$ so we condition on the second coin toss ![]() In this equation, $E=1+EX$, because the tosses are independent, so if the first toss is tails, it is like starting over on the second toss. We first condition on the result of the first coin toss. We solve this problem using a similar approach as in Example 5.6. Let $X$ be the total number of coin tosses. I toss the coin repeatedly until I observe two consecutive heads. \nonumber &=\lambda^2 pq(\lambda^2 pq+ \lambda+1). Join an Oracle Code conference, a series of one-day developer conferences being held worldwide. Learn more about the recent DZone Audience Awards where Java was voted as the favorite programming language. \nonumber &=(\lambda p+ \lambda^2 p^2)(\lambda q+ \lambda^2 q^2)\\ Java Voted as the Favorite Programming Language. To find $P(X \leq 2, Y \leq 4)$, we can write. ![]()
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