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21 Sep 2021, 03:00
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B
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Difficulty:
45%
(medium)
Question Stats:
61% (01:20) correct
39%
(01:57)
wrong
based on 69
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Maya flipped 24 coins, each of which landed with either a head face-up or a tail face up, hereafter referred to as an outcome of “heads” or “tails.”
If all the coins are fair, in that the probability of “heads” = probability of “tails” = \(\frac{1}{2}\) for every toss, which of the following events has the greatest probability?
A. Of the 24 tosses, the number of “heads” equals 11.
B. Of the 24 tosses, the number of “heads” equals 12.
C. Of the 24 tosses, the number of “tails” equals 14.
D. Of the 24 tosses, the number of “tails” is at most 3.
E. Of the 24 tosses, the number of “heads” is it least 22
If all the coins are fair, in that the probability of “heads” = probability of “tails” = \(\frac{1}{2}\) for every toss, which of the following events has the greatest probability?
A. Of the 24 tosses, the number of “heads” equals 11.
B. Of the 24 tosses, the number of “heads” equals 12.
C. Of the 24 tosses, the number of “tails” equals 14.
D. Of the 24 tosses, the number of “tails” is at most 3.
E. Of the 24 tosses, the number of “heads” is it least 22
25 Sep 2021, 12:08
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Bunuel
One of the ways to solve this is:
Since 24 coin is flipped, and probability of “heads” = probability of “tails” = \(\frac{1}{2}\) for every toss
Then highest probability should be getting half heads and half tails
That is 12 head/tail
Option B does that
Hence IMO B
25 Sep 2021, 21:33
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P(Head) = P(Tail) = P(Total Toss)/2
Only condition matches is when P(H) = P(T) = P(24/2) = 12
Corrct Option : B
Posted from my mobile device
Only condition matches is when P(H) = P(T) = P(24/2) = 12
Corrct Option : B
Posted from my mobile device
