Question

A standard poker deck of cards contains 52 cards, of which four are tens. Suppose two cards are drawn sequentially, so that one random circumstance is the result of the first card drawn and the second random circumstance is the result of the second card drawn. Find the probability that the first card is a ten and the second card is not a ten. (Round your answer to four decimal places.)

Answer 1

Answer:

The probability that the first card is a ten and the second card is not a ten is $\frac{16}{221} \ or \ 0.0724$.

Step-by-step explanation:

Consider the provided information.

A standard poker deck of cards contains 52 cards, of which four are tens.

let A is first card is an tens.

B is second card is not an tens.

Now we need to calculate the probability that the first card is a ten and the second card is not a ten.

Which can be calculate as:

$P(A\ and\ B)=P(B|A)\times{P(A)}$

There are 52 card in which four are tens and all cards are equally likely.

Therefore,

$P(A)=\frac{4}{52}$

$P(A)=\frac{1}{13}$

Now let us find the value of $P(B|A)$.

B|A means the second card is not a tens, provided that first card is a tens.

As the first drawn card was tens, that means now we have only 3 tens card in the deck and total 51 cards in the deck.

Now, the total number of cards that are not tens is 51-3 = 48 cards.

So we can the probability of $P(B|A)$ is:

$P(B|A)=\frac{48}{51}$

$P(B|A)=\frac{16}{17}$

Now, substitute the respective values in $P(A\ and\ B)=P(B|A)\times{P(A)}$.

$P(A\ and\ B)=\frac{16}{17}\times{\frac{1}{13}}$

$P(A\ and\ B)=\frac{16}{221}\ or \ 0.0724$

Hence, the probability that the first card is a ten and the second card is not a ten is $\frac{16}{221}\ or \ 0.0724$.