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A fair coin is tossed 4 times. If $X$ a random variable which indicates number of heads, then $\mathrm{P}[\mathrm{X} < 3]=$
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Verified Answer
The correct answer is:
$\frac{11}{16}$
A coin is tossed 4 times
$$
\therefore \mathrm{n}(\mathrm{S})=2^4=16
$$
Following possibilities exist.
(i) All Heads $\Rightarrow 1$ way
(ii) 3 Heads, 1 Tail $\Rightarrow \frac{4 !}{3 !}=4$ ways
(iii) 2 Heads, 2 Tails $\Rightarrow \frac{4 !}{2 ! 2 !}=6$ ways
(iv) 1 Head, 3 Tails $\Rightarrow \frac{4 !}{3 !}=4$ ways
(v) 0 head, 4 Tails $\Rightarrow 1$ way
$\therefore$ Required probability.
$$
\begin{aligned}
& =\mathrm{P}(\mathrm{x}=0,1,2)=\frac{1}{16}+\frac{4}{16}+\frac{6}{16} \\
& =\frac{11}{16}
\end{aligned}
$$
$$
\therefore \mathrm{n}(\mathrm{S})=2^4=16
$$
Following possibilities exist.
(i) All Heads $\Rightarrow 1$ way
(ii) 3 Heads, 1 Tail $\Rightarrow \frac{4 !}{3 !}=4$ ways
(iii) 2 Heads, 2 Tails $\Rightarrow \frac{4 !}{2 ! 2 !}=6$ ways
(iv) 1 Head, 3 Tails $\Rightarrow \frac{4 !}{3 !}=4$ ways
(v) 0 head, 4 Tails $\Rightarrow 1$ way
$\therefore$ Required probability.
$$
\begin{aligned}
& =\mathrm{P}(\mathrm{x}=0,1,2)=\frac{1}{16}+\frac{4}{16}+\frac{6}{16} \\
& =\frac{11}{16}
\end{aligned}
$$
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