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 Let X be a set containing 10 elements and P(X) be its power set.  If A and B are picked up at random from P(X), with replacement, then the probability that A and B have equal number of elements, is :

  • Option 1)

  • Option 2)

    \frac{\left ( 2^{10}-1 \right )}{2^{20}}

  • Option 3)

    \frac{\left ( 2^{10}-1 \right )}{2^{10}}

  • Option 4)


Answers (1)


As learnt in

Probability of occurrence of an event -

Let S be the sample space then the probability of occurrence of an event E is denoted by P(E) and it is defined as 

P\left ( E \right )=\frac{n\left ( E \right )}{n\left ( S \right )}

P\left ( E \right )\leq 1

P(E)=\lim_{n\rightarrow\infty}\left(\frac{r}{n} \right )



- wherein

Where n repeated experiment and E occurs r times.


 Total number of subsets of set X=2^{10}=1024

Number of subsets with one element = ^{10}C_{1}

Number of subsets with two elements= ^{10}C_{2}





Number of subsets with ten elements= ^{10}C_{10}

A and B are taken from P(x) from 2^{10} subsets, total ways = 2^{10}.2^{10}

Number of ways = (^{10}C_{1})^{2}+(^{10}C_{2})^{2}+(^{10}C_{3})^{2}+ ...+(^{10}C_{10})^{2}\:=\: ^{20}C_{10}

Probability = \frac{^{20}C_{10}}{2^{20}}

Option 1)

This option is incorrect.

Option 2)

\frac{\left ( 2^{10}-1 \right )}{2^{20}}

This option is incorrect.

Option 3)

\frac{\left ( 2^{10}-1 \right )}{2^{10}}

This option is incorrect.

Option 4)

This option is correct.

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