Q. 4.     Find the probability distribution of

             (i) number of heads in two tosses of a coin.

Answers (1)

When coin is tossed twice then sample space =\left \{ HH,HT,TH,TT \right \}

Let X be number of heads.

\therefore  X(HH)=2

    X(HT)=1

   X(TH)=1

   X(TT)=0

X can take values of 0,1,2.

P(HH)=P(HT)=P(TH)=P(TT)=\frac{1}{4}

P(X=0)=P(TT)=\frac{1}{4}

P(X=2)=P(HH)=\frac{1}{4}

P(X=1)=P(HT)+P(TH)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}

Table is as shown :

X 0 1 2
P(X) \frac{1}{4} \frac{1}{2} \frac{1}{4}

 

 

 

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