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  • Can someone explain If you drow 4 Playing card Randomone by one without replacement froma well -shuffledpack of 52 cards , what is the probability that they will be of different suits as well asdifferntdenomnations?

If you drow 4 Playing card Random one by one without replacement from a well -shuffled pack  of 52 cards , what is the probability that they will be of different suits as well as differnt denomnations?  

  • Option 1)

    \frac{4}{52C_4}

  • Option 2)

    \frac{4}{52C_2}

  • Option 3)

    \frac{52\times 36\times 24\times 12}{52\times 51\times 50\times 49}

  • Option 4)

    \frac{52\times 36\times 22\times 10}{52\times 51\times 50\times 49}

 

Answers (1)

best_answer

 

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.

 

  Required Probability

=\frac{13_{C_1}\times12_{C_1}\times11_{C_1}\times10_{C_1 }}{52_C_{4}}

=\frac{13_{C_1}\times12_{C_1}\times11_{C_1}\times10_{C_1 }}{52\times 51\times 50 \times 49}

=\frac{13\times12\times11\times10 \times4\times6}{52\times 51\times 50 \times 49}

=\frac{52\times36\times22\times10}{52\times 51\times 50 \times 49}


Option 1)

\frac{4}{52C_4}

Incorrect

Option 2)

\frac{4}{52C_2}

Incorrect

Option 3)

\frac{52\times 36\times 24\times 12}{52\times 51\times 50\times 49}

Incorrect

Option 4)

\frac{52\times 36\times 22\times 10}{52\times 51\times 50\times 49}

Correct

Posted by

prateek

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