excans:Bcc854b3ae: Difference between revisions
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'''Solution: A''' | '''Solution: A''' | ||
We have <math display = "block">P(T>6|T>3) = \frac{P(T>6)}{P(T>3)}</math>. However, <math>T</math> has | We have <math display = "block">P(T>6|T>3) = \frac{P(T>6)}{P(T>3)}</math>. However, <math>T</math> has distribution <math>P(T=k) = (5/6)^{k-1}(1/6) </math> which means that | ||
<math display = "block"> | <math display = "block"> | ||
\frac{P(T>6)}{P(T>3)} = \frac{\sum_{k\geq 7} (5/6)^{k-1}}{\sum_{k \geq 4} (5/6)^{k-1}} = \frac{(5/6)^6}{(5/6)^3} = (5/6)^3 = 0.5787. | \frac{P(T>6)}{P(T>3)} = \frac{\sum_{k\geq 7} (5/6)^{k-1}}{\sum_{k \geq 4} (5/6)^{k-1}} = \frac{(5/6)^6}{(5/6)^3} = (5/6)^3 = 0.5787. | ||
</math> | </math> | ||
<div class="text-center mt-5"> | |||
<youtube>rblPU1vcxj8</youtube> | |||
</div> | |||
Latest revision as of 20:47, 31 March 2025
Solution: A
We have
[[math]]P(T\gt6|T\gt3) = \frac{P(T\gt6)}{P(T\gt3)}[[/math]]
. However, [math]T[/math] has distribution [math]P(T=k) = (5/6)^{k-1}(1/6) [/math] which means that
[[math]]
\frac{P(T\gt6)}{P(T\gt3)} = \frac{\sum_{k\geq 7} (5/6)^{k-1}}{\sum_{k \geq 4} (5/6)^{k-1}} = \frac{(5/6)^6}{(5/6)^3} = (5/6)^3 = 0.5787.
[[/math]]