A card is drawn at random from a standard deck. That card is not put back in the deck, and a second card is drawn at random from the remaining cards in the deck. Neither of the cards drawn so far are put back in the deck, and a third card is drawn at random from the remaining cards in the deck. What is the probability that the first card drawn is a spade, the second card drawn is a heart, and the third card drawn is a heart? Do not round your intermediate computations. Round your final answer to four decimal places.​

The probability of the given scenario occuring is about 0.0153.

Step-by-step explanation:

A standard deck contains 52 cards.

We want the first card to be a spade. In a standard deck, 13 out of the total 52 cards are spades.

So, the probability that the first card is a spade is 13/52 or simply 1/4.

Now, we will draw a second card without replacing the first card. Since we did not replace the first card, the total amount of cards in the deck is now 51.

This time, we want a heart. 13 cards of the remaining 51 will be hearts. So, the probability that the second card is a heart is 13/51.

Now that we've drawn two cards without replacing them, the total number of cards left is 50.

And since we've drawn (or would like to have drawn) a heart as our second card, the total number of cards that are hearts is now 12.

Then the probability of the third card being hearts will be 12/50 or 6/25.

Then the probability that our first card is a spades, second card is a heart, and the third and final card is also a heart without any replacements will be:

The probability of the given scenario occuring is about 0.0153.

true? is that what you're asking or do you mean something else?

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.