Probability..3 cards are drawn from a deck of cards, each replaced and reshuffled..probability of getting?

Question

Im havn trouble with this its for my Stats class...

Suppose that 3 cards are drawn from a deck of cards, each card was replaced and the deck reshuffled. What is the probability of getting one of more hearts?

Answer 1

Since you replace the drawn card each time, the probability of getting a heart for any of the three draws is identical.

There are 13 hearts in a deck of 52 cards, so the probability of getting a heart when drawing a single card is:
13/52 = 1/4 = 0.25

So, the probability of NOT drawing a heart is:
1 - 0.25 = 0.75

The probability of NOT drawing a heart three times (with replacement) is the probability of not drawing a heart to the third power:
(0.75)^3 = 0.4219

So, 42.19% of the time, you will fail to draw a heart in your three draws. That means the probability of drawing one or more hearts is:
1 - 0.4219 = 0.5781 = 57.81%

Answer 2

The chance of getting a heart on a single draw is 1/4, so the probability of not getting one on a single draw is 1-1/4=3/4.
The chance of not getting a heart on any of three draws is (3/4)^3, so the chance of getting at least one is 1-(3/4)^3=37/64.

Answer 3

Hello,

The easiest way is to find the probability of no (0) hearts and subtract from 1. The probability of no hearts is 39P1 / 52P1 = 39/52 so we have this 3 times for each draw so

1 - (39/52 * 39/52 * 39/52) = 1 - (0.75*0.75*0.75) = 1 - .421875 = .578125

Hope This Helps!!

Answer 4

pr 3h) = ( 13/52 )^3 = (1/4)^3 = 1/64

pr (2h) = (1/4)^2 (3/4) x 3= 9/64

pr (h) = 1/4 x (3/4)^2 x 3 = 27 / 64

so add up the above to get the pr of obtaining one of more of the hearts

alternatively find pr ( 0h) and subtract from one

1 - 27/64 = 37/64

(calculator needed here)