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The probability that a certain coin will fall heads up is 50%. What is the probability that it falls heads up on three of six tosses?
(A) \(\frac{41}{2^6}\)
(B) \(\frac{35}{2^6}\)
(C) \(\frac{1}{2^6}\)
(D) \(\frac{21}{2^5}\)
(E) \(\frac{5}{2^4}\)
(A) \(\frac{41}{2^6}\)
(B) \(\frac{35}{2^6}\)
(C) \(\frac{1}{2^6}\)
(D) \(\frac{21}{2^5}\)
(E) \(\frac{5}{2^4}\)
30 Mar 2022, 20:49
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Selecting 3 heads from 6 Outcomes.
6c3/2^6
20/2^6
=5/2^4 (E)
6c3/2^6
20/2^6
=5/2^4 (E)
05 Apr 2022, 04:33
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The probability that the coin shows up a head = 50%; this tells us that it is a fair coin . Therefore, probability of a head showing up = P(H) = \(½\) ; probability of a tail showing up = \(½\).
This fair coin is tossed six times. We need to find out the probability of three heads showing up during these six tosses.
To find the probability of such a compound event, we need to find out the probability of one case and then multiply this with the total number of cases.
One of the cases representing the situation that is desirable looks like {H, H, H, T, T, T}.
Probability of the case above = ½ * ½ * ½ * ½ * ½ * ½ * ½ = \(\frac{1}{2^6}\)
But, three heads can turn up on any of the 6 tosses and hence there are different cases possible; in each of these cases, the probability of 3 heads showing up will be \(\frac{1}{2^6}\) .
Number of different ways in which 3 heads can turn up on any of the 6 tosses = \(\frac{6! }{ 3! * 3!} \)= \(\frac{720 }{ 6 * 6} \)= 20.
{Note that you could just say the number of cases = \(6_C_3\) because effectively, both expressions represent the same number}.
Therefore, probability of 3 heads in 6 tosses of a fair coin = 20 * \(\frac{1}{2^6}\) = 20 * \(\frac{1}{64}\). Upon simplification, we get\( \frac{5}{16}\) which is nothing but\( \frac{5 }{ 2^4}\).
The correct answer option is E.
Answer option C is a very common trap answer marked by those who did not consider the cases and assumed that the situation could only look like {H, H, H, T, T, T}, which is not correct.
This fair coin is tossed six times. We need to find out the probability of three heads showing up during these six tosses.
To find the probability of such a compound event, we need to find out the probability of one case and then multiply this with the total number of cases.
One of the cases representing the situation that is desirable looks like {H, H, H, T, T, T}.
Probability of the case above = ½ * ½ * ½ * ½ * ½ * ½ * ½ = \(\frac{1}{2^6}\)
But, three heads can turn up on any of the 6 tosses and hence there are different cases possible; in each of these cases, the probability of 3 heads showing up will be \(\frac{1}{2^6}\) .
Number of different ways in which 3 heads can turn up on any of the 6 tosses = \(\frac{6! }{ 3! * 3!} \)= \(\frac{720 }{ 6 * 6} \)= 20.
{Note that you could just say the number of cases = \(6_C_3\) because effectively, both expressions represent the same number}.
Therefore, probability of 3 heads in 6 tosses of a fair coin = 20 * \(\frac{1}{2^6}\) = 20 * \(\frac{1}{64}\). Upon simplification, we get\( \frac{5}{16}\) which is nothing but\( \frac{5 }{ 2^4}\).
The correct answer option is E.
Answer option C is a very common trap answer marked by those who did not consider the cases and assumed that the situation could only look like {H, H, H, T, T, T}, which is not correct.
