There are 52 cards in a deck, and 13 of them are hearts. Four cards are dealt, one at a time, off the top of a well-shuffled deck. What is the percent chance that a heart turns up on the fourth card, but not before

10.97%

Step-by-step explanation:

There are 52 cards.

13 of them, are hearts.

Then

52 - 13 = 39 cards are not hearts.

4 cards are drawn, we want to find the percent chance that the fourth card is a heart card, but no before.

So the first card can't be a heart card.

because the deck is well-shuffled, all the cards have the same probability of being drawn.

Then the probability of not getting a heart card, is equal to the quotient between the number of non-heart cards (39) and the total number of cards (52), then the probability is:

p₁ = 39/52

The second card also can't be a heart card, the probability is calculated in the same way than above, but now there are 38 non-heart cards and a total of 51 cards (because one card was already drawn) then the probability here is:

p₂ = 38/51

For the third card the reasoning is similar to the two above cases, here the probability is:

p₃ = 37/50

The fourth card should be a hearts card, the probability is computed in the same way than above, as the quotient between the number of heart cards in the deck (13) and the total number of cards in the deck (now there are 49 cards)

then the probability is:

p₄ = 13/49

The joint probability (the probability of these 4 events happening together) is equal to the product between the individual probabilities:

P = p₁*p₂*p₃*p₄

P = (39/52)*(38/51)*(37/50)*(13/49) = 0.1097

The percent chance is the above number times 100%

Percent =  0.1097*100% = 10.97%

answer is the third choice:

3x^5+4x^4+x^3+2x^2-10x-7=0

step-by-step explanation: