What are the odds in favor of getting at least one head in three successive flips of a coin?
Answer: 1/2
Is this correct?
Could you please explain if I got it wrong?
Thanks.
No. Figure the probability of getting three tails in three tosses, then the probability of getting at least one head is that subtracted from one.
Pr(atleastoneHead)=1-Pr(three tails)
The sample space of outcomes is small enough so we can list them
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
notice that there are 7 with at least one H, and only one with no H
so the odds in favour of getting at least one head in three successive flips of a coin is 7:1
(I hope you meant "odds" and not probability.
The prob of at least one head is 7/8
odds in favour of some some event = (prob of that event):(prob of NOT that event)
Thanks.
Yes, your answer is incorrect. The correct answer to this question is 7/8.
To calculate the odds in favor of getting at least one head in three successive flips of a coin, we can first calculate the probability of not getting a head in any of the flips.
The probability of getting a head in a single flip is 1/2, and the probability of not getting a head (getting a tail) in a single flip is also 1/2. Since the flips are independent events, the probability of not getting a head in all three flips is (1/2) * (1/2) * (1/2) = 1/8.
Since the probability of not getting a head in three flips is 1/8, the probability of getting at least one head in three flips is the complement of that, which is 1 - 1/8 = 7/8.
The odds in favor of getting at least one head in three successive flips of a coin is therefore 7/8.