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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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