Best Answer

We need to determine the separate event.

Let A = obtaining four tails in five flips of coin

Let B = obtaining at least three tails in five flips of coin

Apply Binomial Theorem for this problem, and we have:

P(A | B) = P(A ∩ B) / P(B)

P(A | B) means the probability of "given event B, or if event B occurs, then event A occurs."

P(A ∩ B) means the probability in which both event B and event A occur at a same time.

P(B) means the probability of event B occurs.

Work out each term...

P(B) = (5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0

It's obvious that P(A ∩ B) = (5 choose 4)(½)4(½) since A ∩ B represents events A and B occurring at the same time, so there must be four tails occurring in five flips of coin.

Hence, you should get:

P(A | B) = P(A ∩ B) / P(B)

= ((5 choose 4)(½)4(½))/((5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0)

Study guides

☆☆

Q: What is the probability of obtaining exactly four tails in five flips of a coin if at least three are tails?

Write your answer...

Submit

Still have questions?

Related questions

The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%

The probability of obtaining exactly two heads in three flips of a coin is 0.5x0.5x0.5 (for the probabilities) x3 (for the number of ways it could happen). This is 0.375. However, we are told that at least one is a head, so the probability that we got 3 tails was impossible. This probability is 0.53 or 0.125. To deduct this we need to divide the probability we have by 1-0.125 0.375/(1-0.125) = approximately 0.4286

We can simplify the question by putting it this way: what is the probability that exactly one out of two coin flips is a head? Our options are HH, HT, TH, TT. Two of these four have exactly one head. So 2/4=.5 is the answer.

The requirement that one coin is a head is superfluous and does not matter. The simplified question is "what is the probability of obtaining exactly six heads in seven flips of a coin?"... There are 128 permutations (27) of seven coins, or seven flips of one coin. Of these, there are seven permutations where there are exactly six heads, i.e. where there is only one tail. The probability, then, of tossing six heads in seven coin tosses is 7 in 128, or 0.0546875.

If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8

50%

Pr(3H given >= 2H) = Pr(3H and >= 2H)/Pr(>=2H) = Pr(3H)/Pr(>=2H) = (1/4)/(11/16) = 4/11.

It is 0.6875

If you know that two of the four are already heads, then all you need to find isthe probability of exactly one heads in the last two flips.Number of possible outcomes of one flip of one coin = 2Number of possible outcomes in two flips = 4Number of the four outcomes that include a single heads = 2.Probability of a single heads in the last two flips = 2/4 = 50%.

The probability is 5/16.

Pr(3 flips at least one H) = 1 - Pr(3 flips, NO heads) = 1 - Pr(3 flips, TTT) = 1 - (1/2)3 = 1 - 1/8 = 7/8

11/12

People also asked