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A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.
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Verified Answer
Let p represents the appearance of tail.
$\therefore \quad$ q represents the appearance of head.
Now $\mathrm{q}=3 \mathrm{p} A \mathrm{~s} \mathrm{p}+\mathrm{q}=1 \Rightarrow \mathrm{p}+3 \mathrm{p}=1$
$\Rightarrow p=\frac{1}{4}$ and $q=\frac{3}{4}$
$\mathrm{n}=2($ number of tosses), $r=0,1,2$
$\mathrm{P}(\mathrm{X}=0)={ }^2 C_0 \quad q^2=\left(\frac{3}{4}\right)^2=\frac{9}{16}$
$\mathrm{P}(\mathrm{X}=1)={ }^2 C_1 \quad q p=2 \times \frac{3}{4} \times \frac{1}{4}=\frac{6}{16}$
$\mathrm{P}(\mathrm{X}=2)={ }^2 C_2 p^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}$
Probability distribution
$\therefore \quad$ q represents the appearance of head.
Now $\mathrm{q}=3 \mathrm{p} A \mathrm{~s} \mathrm{p}+\mathrm{q}=1 \Rightarrow \mathrm{p}+3 \mathrm{p}=1$
$\Rightarrow p=\frac{1}{4}$ and $q=\frac{3}{4}$
$\mathrm{n}=2($ number of tosses), $r=0,1,2$
$\mathrm{P}(\mathrm{X}=0)={ }^2 C_0 \quad q^2=\left(\frac{3}{4}\right)^2=\frac{9}{16}$
$\mathrm{P}(\mathrm{X}=1)={ }^2 C_1 \quad q p=2 \times \frac{3}{4} \times \frac{1}{4}=\frac{6}{16}$
$\mathrm{P}(\mathrm{X}=2)={ }^2 C_2 p^2=\left(\frac{1}{4}\right)^2=\frac{1}{16}$
Probability distribution
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