Cards are drawn one by one without replacement from a pack of 52 cards. probability that 10 cards wi

English

Gujarati

Hindi

14.Probability

medium

Cards are drawn one by one without replacement from a pack of $52$ cards. The probability that $10$ cards will precede the first ace is

A

$\frac{{241}}{{1456}}$

B

$\frac{{164}}{{4165}}$

C

$\frac{{451}}{{884}}$

D

None of these

Solution

(b) There are four aces and $48$ other cards.

Therefore the required probability

$ = \frac{{48 \cdot 47 \cdot …. \cdot 39}}{{52 \cdot 51 \cdot …. \cdot 43}}.\frac{4}{{42}} = \frac{{164}}{{4165}}.$

Standard 11

Mathematics

Share

Similar Questions

Two card are drawn successively with replacement from a pack of $52$ cards. The probability of drawing two aces is

easy

Two dice are thrown. The events $A, B$ and $C$ are as follows:

$A:$ getting an even number on the first die.

$B:$ getting an odd number on the first die.

$C:$ getting the sum of the numbers on the dice $\leq 5$

Describe the events $A \cap B^{\prime} \cap C^{\prime}$

easy

The probabilities of a student getting $I, II$ and $III$ division in an examination are respectively $\frac{1}{{10}},\,\frac{3}{5}$ and $\frac{1}{4}.$ The probability that the student fails in the examination is

medium

The probability of choosing at random a number that is divisible by $6$ or $8$ from among $1$ to $90$ is equal to

easy

Two dice are thrown. The events $A, B$ and $C$ are as follows:

$A:$ getting an even number on the first die.

$B:$ getting an odd number on the first die.

$C:$ getting the sum of the numbers on the dice $\leq 5$

Describe the events not $B$

easy