Urijah Lawrence

2023-03-13

From a pack of 52 cards, two cards are drawn in succession one by one without replacement. The probability that both are aces is...

laminimpfv

Beginner2023-03-14Added 3 answers

There are $52$ cards in a pack of cards.

There are $4$ aces in that pack.

Event of choosing one card $n\left(S\right)=C152$

Event of choosing an ace $n\left(A1\right)=C14$

If one card is drawn at random, then the probability of getting an ace is $P\left(A1\right)$

$\Rightarrow $$\frac{n\left(A1\right)}{n\left(S\right)}=\frac{C}{1}$ $\left(\because P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\right)$

$=\frac{4}{52}=\frac{1}{13}$

There are $51$ cards in that pack.

There are $3$ aces in that pack.

Event of choosing one card $n\left(S\right)=C151$

Event of choosing a second ace $n\left(A2\right)=C13$

If one card is drawn at random, then the probability of getting a second ace without any replacement is $P\left(A2\right)$

$\Rightarrow $$\frac{n\left(A2\right)}{n\left(S\right)}=\frac{C}{1}$ ($\because $one card is drawn at random)

$=\frac{3}{51}=\frac{1}{17}$

The odds of receiving two aces without any replacement are calculated by calculating the odds of getting an ace and a second ace without any replacement.

$P\left(A1\right)\times P\left(A2\right)=\frac{1}{13}\times \frac{1}{17}=\frac{1}{221}$

There are $4$ aces in that pack.

Event of choosing one card $n\left(S\right)=C152$

Event of choosing an ace $n\left(A1\right)=C14$

If one card is drawn at random, then the probability of getting an ace is $P\left(A1\right)$

$\Rightarrow $$\frac{n\left(A1\right)}{n\left(S\right)}=\frac{C}{1}$ $\left(\because P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\right)$

$=\frac{4}{52}=\frac{1}{13}$

There are $51$ cards in that pack.

There are $3$ aces in that pack.

Event of choosing one card $n\left(S\right)=C151$

Event of choosing a second ace $n\left(A2\right)=C13$

If one card is drawn at random, then the probability of getting a second ace without any replacement is $P\left(A2\right)$

$\Rightarrow $$\frac{n\left(A2\right)}{n\left(S\right)}=\frac{C}{1}$ ($\because $one card is drawn at random)

$=\frac{3}{51}=\frac{1}{17}$

The odds of receiving two aces without any replacement are calculated by calculating the odds of getting an ace and a second ace without any replacement.

$P\left(A1\right)\times P\left(A2\right)=\frac{1}{13}\times \frac{1}{17}=\frac{1}{221}$

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