Rhoda tosses a coin 5 times. Find the probability of getting 4 heads.
A. 1/32
B. 1/16
C. 3/32
D. 1/8
E. 5/32
The OA is E.
Please, can anyone explain this PS question? I think that the answer would be multiplied by 5 but I'm not sure. I need help. Thanks.
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Rhoda tosses a coin 5 times. Find the probability of getting
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regor60
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The probability of a given pattern of heads or tails in 5 flips is (1/2)^5.
The question asks for probability of getting 4 heads and 1 tail.
Since the 1 tail can occur in any one of the 5 positions, there are 5 patterns of heads and tails.
So 5x(1/2)^5 = [spoiler]E, 5/32[/spoiler]
The question asks for probability of getting 4 heads and 1 tail.
Since the 1 tail can occur in any one of the 5 positions, there are 5 patterns of heads and tails.
So 5x(1/2)^5 = [spoiler]E, 5/32[/spoiler]
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Hi swerve,
We're told that Rhoda tosses a coin 5 times. We're asked for the probability of getting exactly 4 heads. This question can be approached in a couple of different ways, including as a Permutation.
Since each toss has 2 possible outcomes, with 5 tosses there would be (2)^5 = 32 possible arrangements of results. To flip 4 heads, you would also have to flip 1 tail - and it wouldn't be difficult to 'list out' those possibilities:
HHHHT
HHHTH
HHTHH
HTHHH
THHHH
5 outcomes of the 32 possible will give us 4 heads.
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
We're told that Rhoda tosses a coin 5 times. We're asked for the probability of getting exactly 4 heads. This question can be approached in a couple of different ways, including as a Permutation.
Since each toss has 2 possible outcomes, with 5 tosses there would be (2)^5 = 32 possible arrangements of results. To flip 4 heads, you would also have to flip 1 tail - and it wouldn't be difficult to 'list out' those possibilities:
HHHHT
HHHTH
HHTHH
HTHHH
THHHH
5 outcomes of the 32 possible will give us 4 heads.
Final Answer: E
GMAT assassins aren't born, they're made,
Rich
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Jeff@TargetTestPrep
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swerve wrote:Rhoda tosses a coin 5 times. Find the probability of getting 4 heads.
A. 1/32
B. 1/16
C. 3/32
D. 1/8
E. 5/32
The probability of H-H-H-H-T in this order is (½)^4 (½) = 1/32. However, the 4 heads and 1 tail can be arranged in 5!/4! = 5 ways. Therefore, the probability of getting 4 heads and 1 tail is 5 x 1/32 = 5/32.
Answer: E
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