Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET

Previous Year Question (PYQs)

If $X$ is the number of heads obtained in four tosses of a balanced coin. Define $ Y=\dfrac{1}{1+X} $ Find $P\left(Y=\frac{1}{3}\right)$




Solution

$Y=\frac{1}{3} \Rightarrow 1+X=3 \Rightarrow X=2$ So we need: $P(X=2)$ For 4 tosses: $P(X=2)=\binom{4}{2}\left(\frac12\right)^4$ $=\frac{6}{16}=\frac{3}{8}$

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More
Ask Your Question or Put Your Review.

loading...