from this deck is a spade ?......
I think the answer is 1/4, but not sure how they got it...
2006-09-08 02:59:44 · 7 answers · asked by crockett_1969 2 in Education & Reference ➔ Homework Help
It depends what card was lost, either a spade was so there are 12 left, the probability of drawing a spade is 12/51 =.235 or something other than a spade, so the probability is 13/51 =.255.
Now the clincher, if the card that was lost was at RANDOM (good assumption) we can assume 1/4 of the time it would be a spade (13/52 as no cards lost at that point) and 3/4 it wasn't a spade so taking that with the probabilities above, 0.25*12/51+.75*13/51=0.25, the 1/4 you got.
2006-09-08 03:09:07 · answer #1 · answered by Jeremiah W 2 · 0⤊ 0⤋
52 cards in a deck ... one missing is 51 ...
13 cards in a suit 2-10 J Q K A
If missing card is spade 12/51 OR If missing card is not a spade 13/51 so:
p(spade-1) + p(other-1) = 12/51 + 13/51 = 25/51 so I have no idea where 1/4 would come from.
but then again i only got a C in stat so i might be wrong
2006-09-08 03:09:14 · answer #2 · answered by Icon 7 · 0⤊ 0⤋
If the missing card is a spade, then the odds of drawing a spade are 11/51. If the missing card is not a spade, then the odds are 12/51. If you don't know what the missing card is, I don't know how you would calculate it. 1 out of 4 is a good estimate in any case.
2006-09-08 03:03:57 · answer #3 · answered by Random Precision 4 · 0⤊ 0⤋
no the answer is not 1/4, what if the spade itself was the missing card. You have to consider 4 cases and then add up all 4 probabilities as they are mutually exclusive.
2006-09-08 03:08:16 · answer #4 · answered by A 4 · 0⤊ 0⤋
If it is a spade missing and it is the chance that you randomly draw a spade then it is 11/51, if it is the chance that it is not a spade then it is 36/51....but it depends on what it is asking. All the information is not clarified
2006-09-08 03:07:41 · answer #5 · answered by a_streza21 2 · 0⤊ 0⤋
If this is an exceedingly random drawing, and that i assume they are drawn one after the different, then the danger is defined by making use of the probability of having an ace all of the two cases. because of the fact that there are 4 aces in a fifty two card deck, the danger of drawing one is 4/fifty two, or a million/13. Drawing 2 outcomes in the danger of four/fifty two * 4/fifty two, or a million/(13*13) = a million/169
2016-10-14 11:08:56 · answer #6 · answered by ? 4 · 0⤊ 0⤋
13/51. Less than 1/4
2006-09-08 03:06:09 · answer #7 · answered by exel 2 · 0⤊ 0⤋