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Q. 4 Find the probability distribution of 

             (ii) number of tails in the simultaneous tosses of three coins.

Answers (1)

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When 3  coins are  simultaneous tossed  then sample space =\left \{ HHH,HHT,TTH,TTT,HTH,THT,THH,HTT \right \}

Let X be number of tails.

\therefore  X can be 0,1,2,3

  X can take values of 0,1,2.

P(X=0)=P(HHH)=\frac{1}{8}

P(X=1)=P(HHT)+P(HTH)+P(THH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}

P(X=2)=P(THT)+P(HTT)+P(TTH)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}

P(X=3)=P(TTT)=\frac{1}{8}

Table is as shown : 

X 0 1 2 3
P(X) \frac{1}{8} \frac{3}{8} \frac{3}{8} \frac{1}{8}
Posted by

seema garhwal

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