CUET Preparation Today
The probability distribution of number of heads in two tosses of a coin is : |
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The correct answer is Option (1) →
X → no of heads in two tosses $P(X=0)={^2C}_0(\frac{1}{2})^0(\frac{1}{2})^2=\frac{1}{4}$ $P(X=1)={^2C}_1(\frac{1}{2})(\frac{1}{2})=\frac{1}{2}$ $P(X=2)={^2C}_2(\frac{1}{2})(\frac{1}{2})=\frac{1}{4}$ |
