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A Die is Tossed Twice. a 'Success' is Getting an Even Number on a Toss. Find the Variance of Number of Successes. - CBSE (Commerce) Class 12 - Mathematics

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Question

A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes. 

Solution

\[\text{ We have, } \]
\[p = \text{ probability of getting an even number on a toss } = \frac{3}{6} = \frac{1}{2} \text{ and } \]
\[q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}\]
\[\text{ Let X denote a success of getting an even number on a toss . Then } , \]
\[\text{ X follows binomial distribution with parameters n = 2 and }  p = \frac{1}{2}\]
\[ \therefore Var\left( X \right) = \text{ npq }  = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2}\]

  Is there an error in this question or solution?
Solution A Die is Tossed Twice. a 'Success' is Getting an Even Number on a Toss. Find the Variance of Number of Successes. Concept: Random Variables and Its Probability Distributions.
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