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Mathematics, 15.04.2020 04:21 kedjenpierrelouis

Suppose that two cards are randomly selected from a standard 52-card deck. (a) What is the probability that the first card is a clubclub and the second card is a clubclub if the sampling is done without replacement? (b) What is the probability that the first card is a clubclub and the second card is a clubclub if the sampling is done with replacement?

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Answer from: christensenlandon15

(a) $\frac{1}{17}$ (b) $\frac{1}{16}$

Step-by-step explanation:

GIVEN: Suppose that two cards are randomly selected from a standard card deck.

TO FIND: (a) What is the probability that the first card is a club and the second card is a club if the sampling is done without replacement? (b) What is the probability that the first card is a club and the second card is a club if the sampling is done with replacement.

SOLUTION:

(a)

Probability that first card is club $P(A)=\frac{\text{total club cards}}{\text{total cards}}$

$=\frac{13}{52}$

$=\frac{1}{4}$

As sampling is done without replacement.

probability that second card is club $P(B)=\frac{\text{total club cards}}{\text{total cards}}$

$=\frac{12}{51}$

$=\frac{4}{17}$

Probability that first card is club and second card is club $=P(A)\times P(B)$

$=\frac{1}{4}\times\frac{4}{17}=\frac{1}{17}$

(b)

Probability that first card is club $P(A)=\frac{\text{total club cards}}{\text{total cards}}$

$=\frac{13}{52}$

$=\frac{1}{4}$

As sampling is done with replacement.

probability that second card is club $P(B)=\frac{\text{total club cards}}{\text{total cards}}$

$=\frac{13}{52}$

$=\frac{1}{4}$

Probability that first card is club and second card is club $=P(A)\times P(B)$

$=\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}$

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Answer from: Quest

answer: 12 they all have at least 28 days

step-by-step explanation: haha

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Answer from: Quest

16.80 times .20= 3.36

step-by-step explanation:

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