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A two-faced fair coin has its faces designated as head (H) and tail (T). This coin is tossed three times in succession to record the following outcomes: H, H, H. If the coin is tossed one more time, the probability (up to one decimal place) of obtaining H again, given the previous realizations of H, H and H, would be_______
Correct answer is '0.5'. Can you explain this answer?
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A two-faced fair coin has its faces designated as head (H) and tail (T...
Given first three are already heads. If the coin is tossed again, the outcome does not depends on previous outcomes.
Probability of fourth time head is
Given condition is that ( H.H.H) is already realized
Probability of fourth time head is
Given condition is that ( H.H.H) is already realized
Most Upvoted Answer
A two-faced fair coin has its faces designated as head (H) and tail (T...
Solution:
Given, a two-faced fair coin is tossed three times in succession with the outcomes H, H, and H. Now, if the coin is tossed one more time, we need to find the probability of obtaining H again, given the previous realizations of H, H, and H.
The probability of getting H on each toss of the fair coin is 0.5 since the coin has two faces and both are equally likely to appear.
We can approach this problem using the concept of conditional probability. The probability of getting H again, given the previous outcomes of H, H, and H, can be calculated as follows:
P(H|HHH) = P(H and HHH)/P(HHH)
where P(H and HHH) is the probability of getting H on the fourth toss and HHH is the event of getting H on the first three tosses.
The probability of getting H on the fourth toss is 0.5 since each toss of the coin is independent of the previous tosses.
The probability of getting HHH is (0.5)^3 = 0.125 since the coin is fair and the probability of getting H or T on each toss is 0.5.
Therefore,
P(H|HHH) = (0.5)/(0.125) = 4/8 = 0.5
Hence, the probability of obtaining H again, given the previous realizations of H, H, and H, is 0.5.
Given, a two-faced fair coin is tossed three times in succession with the outcomes H, H, and H. Now, if the coin is tossed one more time, we need to find the probability of obtaining H again, given the previous realizations of H, H, and H.
The probability of getting H on each toss of the fair coin is 0.5 since the coin has two faces and both are equally likely to appear.
We can approach this problem using the concept of conditional probability. The probability of getting H again, given the previous outcomes of H, H, and H, can be calculated as follows:
P(H|HHH) = P(H and HHH)/P(HHH)
where P(H and HHH) is the probability of getting H on the fourth toss and HHH is the event of getting H on the first three tosses.
The probability of getting H on the fourth toss is 0.5 since each toss of the coin is independent of the previous tosses.
The probability of getting HHH is (0.5)^3 = 0.125 since the coin is fair and the probability of getting H or T on each toss is 0.5.
Therefore,
P(H|HHH) = (0.5)/(0.125) = 4/8 = 0.5
Hence, the probability of obtaining H again, given the previous realizations of H, H, and H, is 0.5.
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A two-faced fair coin has its faces designated as head (H) and tail (T). This coin is tossed three times in succession to record the following outcomes: H, H, H. If the coin is tossed onemore time, the probability (up to one decimal place) of obtaining H again, given the previous realizations of H, H and H, would be_______Correct answer is '0.5'. Can you explain this answer?
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