In a card game, one of the players wants to draw two DIAMONDS in a row from a newly shuffled deck. (Recall a deck of cards has 52 total cards and 13 of each symbol, including 13 diamonds.)
Create a tree diagram.
What is the probability ONE of the two cards is a diamond?
What is the probability the second card is a diamond, GIVEN the first card drawn was a diamond?
(CLICK YOUR ANSWER FROM THE OPTIONS BELOW. DON'T TYPE YOUR ANSWER IN.)
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Part A

The probability of getting diamonds on the first draw is 13/52 since we have 13 diamonds out of 52 total.

The probability of getting a non-diamond card on the second selection is 39/51 since we have 39 non-diamond cards out of 52-1 = 51 cards left over. Whatever card that was selected first is not put back. In this scenario, we are picking exactly one diamond card.

The probability of both events occurring is (13/52)*(39/51) = 13/68. I skipped a few steps but hopefully you can see how I arrived at that fraction.

The same probability shows up if we had a non-diamond card drawn first followed by a diamond card. This is because we compute (39/52)*(13/51) = 13/68.

So overall, the answer as a fraction is 13/68 + 13/68 = 26/68 = 13/34

The answer in decimal form is roughly 13/34 = 0.3824 which converts to 38.24%

Part B

We're told that "given the first card drawn was a diamond" so we only need to worry about the second card.

We have 13-1 = 12 diamonds left out of 52-1 = 51 total.

The probability of picking another diamond is 12/51 = 0.2353 approximately. This converts to 23.53%

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

: )

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step-by-step explanation: