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A card is drawn from an ordinary deck of 52 well-shuffled

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A card is drawn from an ordinary deck of 52 well-shuffled [#permalink]

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10 Dec 2003, 05:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A card is drawn from an ordinary deck of 52 well-shuffled cards. Find the probability that it is (i) neither red nor an ace; (ii) neither the ten of clubs nor an ace.
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10 Dec 2003, 05:41
prob it is not red is 1/2, prob it is not ace is 2/52( exclude the two red aces ) or 1/52
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10 Dec 2003, 06:14
Prob of not being a ten of clubs is 51/52, prob of not being an ace is 48/52, so the intersection of the events is 51/52x48/52...IMO
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10 Dec 2003, 06:43
favorable outcomes

(1) no reds, only blacks = 26
(2) no reds, only blacks, no aces = 24
(3) no reds, only blacks, no aces, no 10 of clubs = 23

total outcomes = 52

P=23/54
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10 Dec 2003, 07:11
(i) 24/52
(ii) 47/52
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10 Dec 2003, 07:28
I see...

(1) and (2) are not simultaneously applied
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10 Dec 2003, 07:39
Geethu....u r right!!!!
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10 Dec 2003, 08:10
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10 Dec 2003, 08:19
bluefox420 wrote:

(i) neither red nor an ace;
Total Cards =52
No red and no ace = 52-26-2 = 24

P = 24/52

(ii) neither the ten of clubs nor an ace.

No ace and no ten of clubs = 52- 4- 1 =47

P = 47/52
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13 Dec 2003, 00:53
4128851 wrote:
A card is drawn from an ordinary deck of 52 well-shuffled cards. Find the probability that it is (i) neither red nor an ace; (ii) neither the ten of clubs nor an ace.

Quote:
(i) neither red nor an ace;

number of reds = 26
number of aces = 4
number of reds & aces = 2
either red or ace = 28

P( neither red nor ace) = 1 - P(either red or ace)
P(either red or ace) = 28/52
so req prob = 1 - 28/52 = 24/52 = 12/26 = 6/13

Quote:
(ii) Neither the ten of clubs Nor an ace.

# of Ten's of clubs = 1
# of aces = 4
ten's of clubs or aces = 5 cards
total cards = 52

similar method as in 1.
P( either ten of clubs or ace) = 5/52

reqd prob = 1- 47/52
Re: probability   [#permalink] 13 Dec 2003, 00:53
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