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a) a black ace
b) a 4 or 6
c)a red card or 10
d)a red card and a 10
Does it make a difference if it has "or", "and"? If it does, what is the difference?

2007-03-29 09:24:04 · 7 answers · asked by Aggee 1 in Science & Mathematics Mathematics

7 answers

a) 2/52 = 1/26 (ace of clubs & ace of spades)
b) 8/52 = 2/13 (four 4s & four 6s)
c) 28/52 = 7/13 (26 reds plus 10 of clubs & 10 of spades)
d) 2/52 = 1/26 (10 of diamonds & 10 of hearts)

Yes, the use of "or" vs. "and" matters. With 'or', either condition could be satisfied. With 'and', both conditions must be satisfied.

2007-03-29 09:28:02 · answer #1 · answered by Anonymous · 1 0

(a)
there are exactly 2 black aces in a deck, the spade and the club:

P(black ace) = 2 / 52 = 1/26

(b)
There are 4 fours and 6 sixes

P(4 or 6) = P(4) + P(6) - P(4 and 6) = (4/52) + ( 4/52) - 0
= 2 /13

(c)
P(red or 10) = P(red) + P(10) - P(red and 10)
=(26/52) + (4/52) - (2/52) = 28/52

(d)
P(red and 10) = P(a red 10) = 1/26

(e)
OR means the card can be red or a 10 or a red 10. There are lots of cards that are red, but not ten. There are a couple of cards that are 10 but not red.

AND mean strictly a red 10. There are only two of those.

2007-03-29 18:11:23 · answer #2 · answered by Anonymous · 0 0

a) there are two black aces in the deck, so 2/52 = 1/26

b) there are four 4s and 4 6s, so there are 8 chances to "succeed," therefore 8/52 = 2/13

c) The word OR means it could be red or 10 or a red 10. There are 26 red cards, 4 10s BUT 2 of those 10s are red so you are double counting them. So you have 26 + 4 - 2 = 28 chances to get a red or a 10. So the probability is 28/52 = 7/13

d) the word AND mean it has to be BOTH red and 10 and there are two of those cards in the deck (the 10 of hearts and the 10 of diamonds). So again, it's 2/52 = 1/26

2007-03-29 16:37:35 · answer #3 · answered by Kathleen K 7 · 0 0

a) a black ace
2/52 = 1/26

b) a 4 or 6
2*4/52 = 8/52 = 2/13

c) a red card or 10
P(R U 10) = P(R) + P(10) - P(R ∩ 10)
P(R U 10) = 26/52 + 4/52 - 2/52 = 28/52 = 7/13

This is the probability of a red card plus the probability of a 10 minus the probability of a red 10 (which was double counted).

d) a red card and a 10
This is the same as saying a red 10.
2/52 = 1/26

Yes, there is a difference between saying "and" or "or". "And" means both must be true. "Or" means one or both can be true. This is the difference between question c and question d.

2007-03-30 17:08:18 · answer #4 · answered by Northstar 7 · 0 0

a) 2 in 52, or 1 in 26
b) 8 in 52, or 1 in 6.5
c) 1 in 1.86
d) 1 in 26

Yes, or and and make a difference because if it is a red card OR a 10, that means it could be any red card OR it could also be a 10, red or black. If you say a red card AND a 10, it means it has to be a red 10, of which there are only 2 in the deck, so your odds are considerably lower for that happening.

2007-03-29 16:34:09 · answer #5 · answered by fizzygurrl1980 7 · 0 0

a) two black aces in a deck of 52 means that your chances of getting a black ace are 2 out of 52.
b) since there are four 4's and four 6's in a deck, your chances of getting either are 4 out of 52 or 8 out of 52 for both.
c) There are 26 red cards, however there are only four 10s. So your chances of getting a red card are 26 our of 52 or (50%) whereas your chances of getting a 10 are only 4 out 52.
d) Using the example above, only two of the 10s are red, so therefore your chances of getting a red 10 are only 2 out of 52.
"And" means that your chances of getting the correct combination are more restrictive, that is, you are less likely to get the correct combination because the chances of not meeting the condition are higher. That is, you are less likely to get a card that is red and a 10 (2/52). "Or" is mulitplicative, that is you are more likely to get a card that is either a red or a 10, so it increases the chances of the event occurring.

2007-03-29 16:39:29 · answer #6 · answered by phantomlimb7 6 · 0 0

a) 1/26
b) 2/13
c) 7/13
d) i think red and 10 reffers to a red 10. so it is 1/26

2007-03-29 16:38:27 · answer #7 · answered by ganapan7 3 · 0 0

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