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Home  >>  CBSE XII  >>  Math  >>  Probability
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Determine P(E|F) when a coin is tossed three times, where E : head on third toss , F : heads on first two tosses

This is part of a multi-part question in the textbook that is answered separately here.

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Toolbox:
  • In any problem that involves tossing a coin and determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes.
  • Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
The sample space for a coin tossed three times is $S = (HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)$
Total number of outcomes in S = 8.
For our event E: a head occurs on the third toss, $\rightarrow E = (HHH, THH, HTH, TTH)$. Total number of outcomes in E = 4 .
For our event F: heads on the first two tosses $\rightarrow F = (HHH, HHT). The total number of outcomes in F = 2.
$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{4}{8} = \frac{1}{2}$
$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{2}{8} = \frac{1}{4}$
For our event E: a head occurs on the third toss, $\rightarrow E = (HHH, THH, HTH, TTH)$ and F: heads on the first two tosses $\rightarrow F = (HHH, HHT) \rightarrow (E \;\cap\; F) = (HHH)$
For our event (E \;\cap\; F) = (HHH)$. Total number of favorable outcome is 1.
$\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}} = \frac{1}{8}$
Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
$\Rightarrow \large \frac{P(E \;\cap \;F)}{P(F)} $$=\Large \frac {\Large \frac{1}{8}} {\Large \frac{1}{4}}$ = $\large \frac{1}{2}$
answered Jun 18, 2013 by balaji.thirumalai
 

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