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Find the probability distribution of number of heads in four tosses of a coin.

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  • Once we write down the sample space for the problem we can find the number of heads or tails for the favorable outcomes in each case and find their probabilities and arrange as a probability distribution.
Given: 4 tosses of a coin, the sample space can be represented as follows: S = $\begin{Bmatrix} HHHH & HHHT & HHTH &HHTT \\ HTHH &HTHT &HTTH &HTTT \\ THHH & THHT &THTH &THTT \\ TTHH &TTHT &TTTH & TTTT \end{Bmatrix}$
All these events are equally likely and have the same probability = $\large\frac{1}{16}$
Let X be the number of heads in those tosses. We can see that X can take the values of 0,1,2,3 or 4. It follows that:
P (X = 0) = P (TTTT) = $\large\frac{1}{16}$
P (X = 1) = P (HTTT, THTT, TTHT, TTTH) = $\large\frac{4}{16} = \frac{1}{4}$
P (X = 2) = P (HHTT, HTHT, HTTH, TTHH, THHT, THTH) = $\large\frac{6}{16} = \frac{3}{8}$
P (X = 3) = P (HHHT, HHTH, HTHH, THHH) = $\large\frac{4}{16} = \frac{1}{4}$
P (X = 4) = P (HHHH) = $\large\frac{1}{16}$
Therefore we can write the probability distribution as follows:
$\begin{matrix} \textbf{X} & 0 &1 &2&3&4 \\ \textbf{P(X)} &\large \frac{1}{16} &\large \frac{1}{4} &\large \frac{3}{8} &\large\frac{1}{4}&\large\frac{1}{16}\end{matrix}$
answered Jun 19, 2013 by balaji.thirumalai

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