excans:0bd71f391d: Difference between revisions

Latest revision as of 15:15, 26 June 2024

Solution: C

Let [math]T[/math] equal the number of tosses until the first head appears. [math]T[/math] has distribution [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}{1-0.5-0.25} = \frac{(1/2)^5}{(1/2)^2} = (1/2)^3 = 0.125. [[/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 distribution <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 } = \frac{(1/2)^5}{(1/2)^2} = (1/2)^3 = 0.125.
+P(T=5|T>2) = \frac{P(T=5)}{P(T>2)} = \frac{(1/2)^5}{1-0.5-0.25} = \frac{(1/2)^5}{(1/2)^2} = (1/2)^3 = 0.125.
 </math>