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Q. 7 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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the coin is tossed twice, total outcomes =4 $=\left \{ HH,TT,HT,TH \right \}$

probability of getting a tail be x.

i.e. $P(T)=x$

Then $P(H)=3x$

$P(T)+P(H)=x+3x=1$

$4x=1$

$x=\frac{1}{4}$

$P(T)=\frac{1}{4}$ and $P(H)=\frac{3}{4}$

Let X : number of tails

No tail : $P(X=0)=P(H).P(H)=\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$

1 tail : $P(X=1)=P(HT)+P(TH)=\frac{3}{4}\times \frac{1}{4}+\frac{1}{4}\times \frac{3}{4}=\frac{3}{8}$

2 tail : $P(X=2)=P(TT)=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

the probability distribution of number of tails are

X	0	1	2
P(X)	$\frac{9}{16}$	$\frac{3}{8}$	$\frac{1}{16}$

Posted by

seema garhwal

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