from a srandard deck of 52 cards, what is the probability of drawing two five cards with replacement?

2 answers:

4 0

Event A = drawing a card with a "5" on the label.

There are 4 ways that event A could occur: drawing a '5' of hearts, 5 of spades, 5 of diamonds, or five of clubs.

This is out of 52 cards total

For one card,
P(A) = 4/52 = 1/13

For two cards,
P(A and A) = P(A)*P(A)
P(A and A) = (1/13)*(1/13)
P(A and A) = 1/169
We're able to multiply like this because the events are independent (replacement occurs).

The answer as a fraction is 1/169
The answer in decimal form is approximately 0.005917
The answer as a percent is approximately 0.5917%

Alika [10]2 years ago

3 0

There are 4 5's in a deck of 52 cards so
P(5) = 4/52
Since the card is being replaced it will be the same probability the second time.
P(5) = 4/52
P(5)= 4/52 * P(5)=4/52 = 16/2704 = 1/169

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For each $n \in \mathbb{N}$, let $A_n = [n] \times [n]$. Define $B = \bigcup_{n \in \mathbb{N}} A_n$. Does $B = \mathbb{N} \time

andre [41]

Answer:

No, it is not.

Step-by-step explanation:

The set $C = \mathbb{N} \times \mathbb{N}$ contains every ordered pair of Natural numbers, while B only contains those pairs in which both values in each entry are the same. Therefore, C is a bigger set than B, but B is not equal to C because for example C contains $[1] \times [2]$ and B doesnt because 1 is not equal to 2.