Subject

    Quantitative Aptitude

    Topic

    Probability

    A fair coin is tossed 10 times. What is the probability that only the first two tosses will yield tails?

    ক)
    \({1 \over 2}\)
    খ)
    \({\left( {{1 \over 2}} \right)^2}\)
    গ)
    \({}^{10}{C_2}{\left( {\frac{1}{2}} \right)^2}\)
    ঘ)
    \({\left( {\frac{1}{2}} \right)^{10}}\)
    ঙ)
    \({\left( {\frac{1}{2}} \right)^8}\)

    Explanation

    P(E)=Number of favorable outcomesNumeber of possible outcomes=n(E)n(S)P\left( E \right) = \frac{{Number\ of\ favorable\ outcomes}}{{Numeber\ of\ possible\ outcomes}} = \frac{{n\left( E \right)}}{{n\left( S \right)}} ⇒ Probability of getting tail in one coin = ½, ⇒ Probability of not getting tail in one coin = 1- ½ = ½, Hence, All the ten tosses are independent of each other. ∴ Required probability =(12)2×(12)8=(12)10= {\left( {\frac{1}{2}} \right)^2} \times {\left( {\frac{1}{2}} \right)^8} = {\left( {\frac{1}{2}} \right)^{10}}

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