excans:0bd71f391d: Difference between revisions
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(Created page with "Let <math>T</math> equal the number of tosses until the first head appears. <math>T</math> has a geometric distribution with <math>P(T=k) = (1/2)^k </math> for <math>k\geq 1 </math>. The probability that the first head appears on the fifth toss, given that it has not appeared in the first two tosses, equals <math display = "block"> P(T=5|T>2) = \frac{P(T=5)}{P(T>2)} = \frac{(1/2)^5}{\sum_{k \geq 3} (1/2)^k } </math>") |
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'''Solution: C''' | |||
Let <math>T</math> equal the number of tosses until the first head appears. <math>T</math> has a geometric distribution with <math>P(T=k) = (1/2)^k </math> for <math>k\geq 1 </math>. The probability that the first head appears on the fifth toss, given that it has not appeared in the first two tosses, equals | Let <math>T</math> equal the number of tosses until the first head appears. <math>T</math> has a geometric distribution with <math>P(T=k) = (1/2)^k </math> for <math>k\geq 1 </math>. The probability that the first head appears on the fifth toss, given that it has not appeared in the first two tosses, equals | ||
<math display = "block"> | <math display = "block"> | ||
P(T=5|T>2) = \frac{P(T=5)}{P(T>2)} = \frac{(1/2)^5}{\sum_{k \geq 3} (1/2)^k } | P(T=5|T>2) = \frac{P(T=5)}{P(T>2)} = \frac{(1/2)^5}{\sum_{k \geq 3} (1/2)^k } = \frac{(1/2)^5}{(1/2)^2} = (1/2)^3 = 0.125. | ||
</math> | </math> | ||
Revision as of 15:07, 26 June 2024
Solution: C
Let [math]T[/math] equal the number of tosses until the first head appears. [math]T[/math] has a geometric distribution with [math]P(T=k) = (1/2)^k [/math] for [math]k\geq 1 [/math]. The probability that the first head appears on the fifth toss, given that it has not appeared in the first two tosses, equals
[[math]]
P(T=5|T\gt2) = \frac{P(T=5)}{P(T\gt2)} = \frac{(1/2)^5}{\sum_{k \geq 3} (1/2)^k } = \frac{(1/2)^5}{(1/2)^2} = (1/2)^3 = 0.125.
[[/math]]