Question

Two cards are drawn simultaneously from a well-shuffled pack of 52 cards. If X is the random variable of getting queens, then the value of 2E(X) + 3E(X²) for the number of queens is

पर्याय

• \[\frac{132}{221}\]

• \[\frac{108}{221}\]

• \[\frac{176}{221}\]

• \[\frac{68}{221}\]

Answer 1

\[\frac{176}{221}\]

Explanation:

Let X denote the number of queens in a draw of two cards.
X can assume values 0, 1, 2

P(X = 0) = P(no queen)

\[=\frac{^{48}\mathrm{C}_2}{^{52}\mathrm{C}_2}=\frac{48\times47}{52\times51}=\frac{188}{221}\]

P(X = 1) = P(exactly one queen)

\[=\frac{^4\mathrm{C}_1\times^{48}\mathrm{C}_1}{^{52}\mathrm{C}_2}=\frac{32}{221}\]

P(X = 2) = P(both queens)

\[=\frac{^4\mathrm{C}_2}{^{52}\mathrm{C}_2}=\frac{1}{221}\]

 E(X) = \[=0\times\frac{188}{221}+1\times\frac{32}{221}+2\times\frac{1}{221}\]

\[=\frac{34}{221}\]

 E(X²) = \[=0\times\frac{188}{221}+1\times\frac{32}{221}+4\times\frac{1}{221}\]

\[=\frac{36}{221}\]

\[2\mathrm{E}(\mathrm{X})+3\mathrm{E}(\mathrm{X}^2)\] \[=2\times\frac{34}{221}+3\times\frac{36}{221}\]

\[=\frac{176}{221}\]