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# If E is defined to be drawing a black card out of a normal 52 card dec

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Manager
Joined: 26 Dec 2018
Posts: 142
Location: India
If E is defined to be drawing a black card out of a normal 52 card dec  [#permalink]

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10 Jan 2019, 03:07
3
00:00

Difficulty:

25% (medium)

Question Stats:

69% (00:55) correct 31% (01:18) wrong based on 31 sessions

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If E is defined to be drawing a black card out of a normal 52 card deck and F is defined to be drawing a heart, what is the probability of either E or F coming true?

A) 1/2
B) 2/3
C) 1
D) 5/8
E) 3/4

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Intern
Joined: 09 Nov 2018
Posts: 2
Re: If E is defined to be drawing a black card out of a normal 52 card dec  [#permalink]

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10 Jan 2019, 03:56
P(A)= 1/2
P(B)=1/4
P(A) or P(B)= 1/2+1/4-1/2*1/4= 5/8

Ans D

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Manager
Joined: 26 Dec 2018
Posts: 142
Location: India
Re: If E is defined to be drawing a black card out of a normal 52 card dec  [#permalink]

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11 Jan 2019, 12:10
PlatinumGMAT OFFICIAL EXPLANATION :

The question is asking for P(E∪F), which is technically referred to as the union of events E and F (i.e., the probability of either event E or event F coming true).
The statistical formula for this type of problem is:
P(E∪F) = P(E) + P(F) - P(E∩F)
In words: the probability of events E or F = the probability of event E + the probability of event F - the probability of both events E and F occurring simultaneously
Note: It is necessary to subtract P(E∩F) in order to avoid double counting.
P(E) = 1/2 since all cards are either black or red and, as a result, half of the cards are black.
P(F) = 1/4 since there are four types of cards (i.e., hearts, diamonds, clubs, or spades).
P(E∩F) = 0 since it is impossible to pick a card that is both black and a heart because of the fact that hearts are red cards (not black cards).
P(E∪F) = P(E) + P(F) - P(E∩F)
P(E∪F) = (1/2) + (1/4) - 0 = 3/4

For students with a more advanced understanding of statistics, you will notice that E and F are mutually exclusive. As a result, a simpler statistical formula can be used:
E and F Mutually Exclusive
Determine if E and F are mutually exclusive.
P(E∩F) = 0. It is impossible to draw a card that is both black and a heart. Thus, the two events are mutually exclusive.
P(E∪F) = P(E) + P(F)
P(E) = 1/2 as half of a deck is composed of black cards while the other half is composed of red cards.
P(F) = 1/4 as one fourth of the deck is composed of hearts.
P(E∪F) = P(E) + P(F) = 1/2 + 1/4 = 2/4 + 1/4 = 3/4
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Re: If E is defined to be drawing a black card out of a normal 52 card dec   [#permalink] 11 Jan 2019, 12:10
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