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INVERSE TRIGNOMETRIC FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINVERSE TRIGNOMETRIC FUNCTIONS5 Marks
Question
Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the number of kings.
✓

Answer

Let X denote the number of kings obtained in two draws. X takes the values 0, 1, 2.
Let p denote probability of first king.
$\therefore \ \text{p}=\frac{4}{52},\text{q}=\frac{48}{52}$
$\text{P}(\text{X}=0)=\frac{48}{52}\times\frac{47}{51}=\frac{12}{13}\times\frac{47}{51}=\frac{188}{221}$
$\text{P}(\text{X}=1)=\frac{4}{52}\times\frac{48}{51}+\frac{48}{52}\times\frac{4}{51}$
$=\frac{16}{221}+\frac{16}{221}=\frac{32}{221}$
$\text{P}(\text{X}=2)=\frac{4}{52}\times\frac{3}{51}$
$=\frac{1}{13}\times\frac{1}{17}$
$=\frac{1}{221}$
$\therefore$ probability distribution is:
X 0 1 2
P(X) $\frac{188}{221}$ $\frac{32}{221}$ $\frac{1}{221}$
Now,
Mean of $\text{X}=\text{E}(\text{X})=\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_{\text{i}}{\text{ p(x}_{\text{i}}})$
$=0\times\frac{188}{221}+1\times\frac{32}{221}+2\times\frac{1}{221}$
$0+\frac{32}{221}+\frac{2}{221}=\frac{34}{221}$
Also, $\text{E}(\text{X})^2=\sum\limits^{\text{n}}_{\text{i}=1}\text{x}_{\text{i}}^2{\text{ p(x}_{\text{i}}})$
$=0^2\times\frac{188}{121}+1^2\times\frac{32}{221}+2^2\times\frac{1}{221}$
$=0+\frac{32}{221}+\frac{4}{221}=\frac{36}{221}$
$\therefore \ \text{var}(\text{X})=\text{E}(\text{X})^2-[\text{E}(\text{X})]^2$
$=\frac{36}{221}-\Big(\frac{34}{221}\Big)^2=\frac{36}{221}-\frac{1156}{48841}$
$=\frac{7956-1156}{48841}$
$=\frac{6800}{48841}$
$\therefore\sigma^2=\sqrt{\text{var}(\text{X})}=\frac{\sqrt{6800}}{221}=\frac{82.46}{221}=0.37$

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