# Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Let X denote the random variable of the number of aces obtained in the two drawn cards. Then $P(X= 1) + P(X = 2)$ equals:Option 1)$\frac{49}{169}$Option 2)$\frac{52}{169}$Option 3)$\frac{24}{169}$Option 4)$\frac{25}{169}$

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

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

$P\left ( E \right )\leq 1$

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

In a pack of card

4 Ace's and 48 non-Ace's

$P\left ( x=1 \right )=\frac{4c_{1}}{^{52}\textrm{C}_{1}}\times \frac{^{48}\textrm{C}_{1}}{^{52}\textrm{C}_{1}}+\frac{^{48}\textrm{C}_{1}}{^{52}\textrm{C}_{1}}\times \frac{^{4}\textrm{C}_{1}}{^{52}\textrm{C}_{1}}=\frac{24}{169}$

$P\left ( x=2 \right )=\frac{^{4}\textrm{C}_{1}}{^{52}\textrm{C}_{1}}\times \frac{^{4}\textrm{C}_{1}}{^{52}\textrm{C}_{1}}=\frac{1}{169}$

$P\left ( x=1 \right )+P\left ( x=2 \right )=\frac{25}{169}$

- wherein

Where n repeated experiment and E occurs r times.

Option 1)

$\frac{49}{169}$

Option 2)

$\frac{52}{169}$

Option 3)

$\frac{24}{169}$

Option 4)

$\frac{25}{169}$

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