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A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is

Option: 1

\frac{1}{8}


Option: 2

\frac{1}{2}


Option: 3

\frac{3}{8}


Option: 4

\frac{3}{4}


Answers (1)

best_answer

Let \mathrm{E_1}  is the event getting head in first toss and  \mathrm{E_2}  is the event, getting exactly two heads in three tosses then


\mathrm{P\left(E_2 / E_1\right) =\frac{P\left(E_2 \cap E_1\right)}{P\left(E_1\right)} P\left(E_2 \cap E_1\right) =P(H H T, H T H)}
\mathrm{ =\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4} }
and     \mathrm{P\left(E_1\right) =\frac{1}{2}}    

\mathrm{\therefore \quad P\left(E_2 / E_1\right)=\frac{1 / 4}{1 / 2}=\frac{1}{2}}.
 

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manish

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