Find the probability distribution of the number of heads in two tosses of a coin.
(N/A) When one coin is tossed twice,the sample space is $S = \{HH, HT, TH, TT\}$.
Let $X$ represent the number of heads.
Then,$X(HH) = 2, X(HT) = 1, X(TH) = 1, X(TT) = 0$.
Therefore,$X$ can take the values $0, 1,$ or $2$.
It is known that $P(HH) = P(HT) = P(TH) = P(TT) = \frac{1}{4}$.
$P(X=0) = P(TT) = \frac{1}{4}$.
$P(X=1) = P(HT) + P(TH) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
$P(X=2) = P(HH) = \frac{1}{4}$.
Thus,the required probability distribution is as follows:
Let $X$ represent the number of heads.
Then,$X(HH) = 2, X(HT) = 1, X(TH) = 1, X(TT) = 0$.
Therefore,$X$ can take the values $0, 1,$ or $2$.
It is known that $P(HH) = P(HT) = P(TH) = P(TT) = \frac{1}{4}$.
$P(X=0) = P(TT) = \frac{1}{4}$.
$P(X=1) = P(HT) + P(TH) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
$P(X=2) = P(HH) = \frac{1}{4}$.
Thus,the required probability distribution is as follows:
| $X$ | $0$ | $1$ | $2$ |
| $P(X)$ | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ |
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