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Question

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$ \begin{align}

& \text{a) }\dfrac{2}{5} \\

& \text{b) }\dfrac{3}{5} \\

& \text{c) }\dfrac{1}{2} \\

& \text{d) }\dfrac{17}{40} \\

\end{align} $

Answer
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Now it is given that a coin is tossed 40 times and among the 40 times it showed tail 24 times

Now the result of a coin toss can either be head or tail hence if it showed tail 24 times means that the remaining outcomes were heads.

This means out of 40 outcomes, If 24 outcomes were tail then 40 – 24 outcomes were head.

Hence, 16 outcomes were head. ……(1)

Now let B be the event such that we get ahead. Then from equation (1) we get

$ n(B)=16.............(2) $

Now we know that the formula of probability is $ \dfrac{\text{number of required outcomes}}{\text{Total outcomes}} $

Now from equation (2) we know the number of required outcomes is 16.

Also the total outcome is 40

Hence applying these formula we get $ p(B)=\dfrac{16}{40} $

Let us reduce the fraction by dividing the numerator and denominator by 8.

Hence, $ p(B)=\dfrac{2}{5} $

Hence we get the probability of getting a head is $ \dfrac{2}{5} $

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