Hailey L. 2 Answers By Expert Tutors
James C. answered • 09/17/19 Doctoral Candidate (and a Master's) in Statistics With Nine Years Exp.
The probability of choosing a heart, P(Heart) = 13/52 = 0.25. Without replacement, you now have 51 cards left in the deck. So the probability of subsequently choosing a Spade is, P(Spade) = 13/51. The probability of choosing a heart AND then a space, without replacement is: (13/52) * (13/51) = .06372549 (Remember, "AND" statements envoke the multiplication rule; "OR" statements envoke the additional rule. And when you sample WITH replacement, you have to then account for the missing cards in subsequent draws.) I hope this helps. James
Nestor R. answered • 09/17/19 Professional experienced using mathematics and statistics
Each deck of 52 cards has 13 hearts, 13 clubs, 13 spades and 13 diamonds. For the first draw, the probability of a heart is 13/52 = 1/4 = 0.2500. The card is not replaced, so 51 cards remain in the deck. For the second draw, the probability of a spade is 13/51 = 0.2549. Thus the probability of drawing a heart then a spade, without repacement, is 13/52*13/51 = 0.2500*0.2549 = 0.0637 $\begingroup$
Suppose you draw two cards from a deck of 52 cards without replacement. 1) What is the probability that both cards are hearts? 2) What is the probability that exactly one of the cards is hearts? 3) What is the probability that none of the cards are hearts? I get that the first answer is 13/52 * 12/51. For the second, is it 13/52 * 39/51 because the new deck has 51 cards of which 12 are hearts, hence 39/51 is the probability of the second not being hearts? Will the third answer just be 39/52 * 38/51? $\endgroup$ There are 13 hearts in the deck of 52. The probability of getting the first heart is 13/52=1/4. The probability of getting the second heart is 12/51. Since they're independent events, the probability of two hearts is . . . Numbers less than 8 are Ace, 2, 3, 4, 5, 6, 7, which are 7 cards in 4 suits or 28 total.
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