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}

Answers (1)
A admin

 

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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