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10 Dec 2021, 03:15
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Difficulty:
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Question Stats:
37% (02:18) correct
63%
(02:22)
wrong
based on 52
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A bag contains 65 coins. 64 of the coins are fair but 1 coin contains two heads. If a coin, chosen at random from the bag and then tossed, turns up heads 6 times in a row, what is the probability that it is the two-headed coin?
A. 1/64
B. 2/65
C. 1/6
D. 1/2
E. 5/8
Are You Up For the Challenge: 700 Level Questions
A. 1/64
B. 2/65
C. 1/6
D. 1/2
E. 5/8
Are You Up For the Challenge: 700 Level Questions
10 Dec 2021, 05:44
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Probability of selecting a fair coin:
64/65
Probability of this coin coming up heads 6 times:
(1/2)^6 = 1/64
Probability of both events occurring:
(64/65)*(1/64) = 1/65
Probability of unfair coin being selected:
1/65
Probability of this coin coming up heads:
1
Probability of both events occurring:
(1/65)*1 = 1/65
Total probability of 6 heads in a row:
(1/65)+(1/65) = 2/65
Probability of the unfair coin having been selected:
(1/65)/(2/65) = 1/2
Posted from my mobile device
64/65
Probability of this coin coming up heads 6 times:
(1/2)^6 = 1/64
Probability of both events occurring:
(64/65)*(1/64) = 1/65
Probability of unfair coin being selected:
1/65
Probability of this coin coming up heads:
1
Probability of both events occurring:
(1/65)*1 = 1/65
Total probability of 6 heads in a row:
(1/65)+(1/65) = 2/65
Probability of the unfair coin having been selected:
(1/65)/(2/65) = 1/2
Posted from my mobile device
29 Feb 2024, 03:06
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To solve this problem, we can use Bayes' theorem to calculate the probability that the chosen coin is the two-headed coin, given that it turned up heads 6 times in a row.
Let's define the following events:
A: The chosen coin is the two-headed coin.
B: The coin turned up heads 6 times in a row.
We need to find P(A|B), which is the probability that the chosen coin is the two-headed coin, given that it turned up heads 6 times in a row.
Using Bayes' theorem, we have:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) is the probability that the coin turns up heads 6 times in a row given that it is the two-headed coin. Since the two-headed coin always lands on heads, P(B|A) = 1.
P(A) is the probability of choosing the two-headed coin from the bag. There are 65 coins in total, and only 1 of them is the two-headed coin. Therefore, P(A) = 1/65.
P(B) is the probability that the coin turns up heads 6 times in a row. This can happen in two ways: either we choose the two-headed coin and it lands on heads 6 times in a row (probability 1), or we choose one of the fair coins and it lands on heads 6 times in a row (probability (1/2)^6 = 1/64).
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
= 1 * (1/65) + (1/64) * (64/65)
= 1/65 + 1/65
= 2/65
Now we can calculate P(A|B):
P(A|B) = (P(B|A) * P(A)) / P(B)
= (1 * (1/65)) / (2/65)
= 1/2
Answer is D
Let's define the following events:
A: The chosen coin is the two-headed coin.
B: The coin turned up heads 6 times in a row.
We need to find P(A|B), which is the probability that the chosen coin is the two-headed coin, given that it turned up heads 6 times in a row.
Using Bayes' theorem, we have:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) is the probability that the coin turns up heads 6 times in a row given that it is the two-headed coin. Since the two-headed coin always lands on heads, P(B|A) = 1.
P(A) is the probability of choosing the two-headed coin from the bag. There are 65 coins in total, and only 1 of them is the two-headed coin. Therefore, P(A) = 1/65.
P(B) is the probability that the coin turns up heads 6 times in a row. This can happen in two ways: either we choose the two-headed coin and it lands on heads 6 times in a row (probability 1), or we choose one of the fair coins and it lands on heads 6 times in a row (probability (1/2)^6 = 1/64).
P(B) = P(B|A) * P(A) + P(B|not A) * P(not A)
= 1 * (1/65) + (1/64) * (64/65)
= 1/65 + 1/65
= 2/65
Now we can calculate P(A|B):
P(A|B) = (P(B|A) * P(A)) / P(B)
= (1 * (1/65)) / (2/65)
= 1/2
Answer is D
