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Find the mean number of heads in three tosses of a fair coin.
CBSE | Class 10 | Guides | Maths | NCERT | Probability
Find the mean number of heads in three tosses of a fair coin.

Solution:

Three times, a coin is tossed.

Three coins are flung at the same time. As a result, the experiment’s sample space is \mathrm{S}= \{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT\}. The number of heads is represented by \mathrm{X}.

As we can see, \mathrm{X} is a sample space function with a range of \{0,1,2,3\}.

As a result, \mathrm{X} is a random variable with values of 0,1,2, or 3.

P(X=0)=P(T T T)=1 / 8

\mathrm{P}(\mathrm{X}=1)=\mathrm{P}(\mathrm{TTH})+\mathrm{P}(\mathrm{THT})+\mathrm{P}(\mathrm{HTT})=1 / 8+1 / 8+1 / 8=3 / 8

P(X=2)=P(T H H)+P(H T H)+P(H H T)=1 / 8+1 / 8+1 / 8=3 / 8

P(X=3)=P(H H H)=1 / 8

As a result, the necessary probability distribution is,

Therefore mean \mu is:

\mu=E(X)=\sum_{i=1}^{n} x_{i} p_{i}

We may now retrieve the result by swapping the values.

=0 \times \frac{1}{8}+1 \times \frac{3}{8}+2 \times \frac{3}{8}+3 \times \frac{1}{8}

=\frac{3}{8}+\frac{3}{4}+\frac{3}{8}=\frac{3+6+3}{8}=\frac{12}{8}

\Rightarrow \mu=E(X)=\frac{3}{2}=1.5

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