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Q: What is the probability of getting a least one tail when tossing eight fair coins?

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It is 0.375

One hundred percent it you toss the coin eight times.

Each time you toss the die the probability of rolling an even number is 3 out of 6 or 1/2. So, the probability of tossing three consecutive even numbers is (1/2)3 = 1/8 = 0.125, which is one chance in eight.

the answer is 2/13

There are eight possible outcomes: HHH, HHT, HTT, HTH, TTT, TTH, THH, THT. Of these, 3 contain two tails: HTT, TTH, and THT and the probability of getting two tails is 3/8. If the question were 'getting at least two tails' then TTT would need to be included for a probability of 4/8 or 0.5.

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Is possible.

It is 0.375

One hundred percent it you toss the coin eight times.

Each time you toss the die the probability of rolling an even number is 3 out of 6 or 1/2. So, the probability of tossing three consecutive even numbers is (1/2)3 = 1/8 = 0.125, which is one chance in eight.

the answer is 2/13

Probability that it is one of these eight cards is 8/52. Hence the probability of not getting these eight cards is 44/52

The probability is 61.538 if you are starting with a fresh deck and it is the first draw.

There are eight possible outcomes: HHH, HHT, HTT, HTH, TTT, TTH, THH, THT. Of these, 3 contain two tails: HTT, TTH, and THT and the probability of getting two tails is 3/8. If the question were 'getting at least two tails' then TTT would need to be included for a probability of 4/8 or 0.5.

The probability of getting exactly seven tails if you flip a coin eight times is: P(7T1H) = 8∙(1/2)8 =0.03125 ≈ 3.1%

One in eight, or 12.5%.

The probability of getting 8 heads out of 10 tosses is (10C8)(1/2)^8 (1/2)^2 = 45 / 1024 = 0.0439. It is assumed that the probability of getting a head in a single toss is 1/2. 10C8 = 10 x 9 / 1 x 2 = 45

Looking at the total possibilities, you have eight different outcomes: TTT TTH THH HHH HHT HTT HTH THT Counting your two heads, one tails, you get a total of 3 possibilities out of 8, or 3/7

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