The probability to win a gift

Question

Everyone has 5% chances to win a gift then how about the chance of a 4 men's team to win 1 gift? Is 0.95^4 right? How to calculate it with 5% ?

Answer 1

The odds of a 4 man team is similar to the odds of just one person playing this game 4 times, with 5% chance each time.

If any of the 4 win, the team wins. If the one person wins on any of the 4 tries, they win, although the one person may win 1 to 4 gifts, while the team only wins 1 gift. Still very similar

a 5% chance of winning each time, and asking will s/he win at least once, is the same as asking what is the chance of not winning at all, then subtract that from 100%

If you have a 95% chance of losing each time then the only way you lose 4 times is 0.95 to the 4th power

(.95)⁴ = .81.5 Odds of winning are 1- odds of losing or 1-81.5 = 0.185 = 18.,5% chance of winning

Maybe the 4 person team has a similar chance of a win, although it will only win one gift while the single person could win 1 to 4 gifts.

Answer 2

The probability for a person to win is independent so this calls for the binomial probability distribution:

size of the set = n = 4

desired successes = k = 1

probability of success = p = 0.05

probability of failure = q = 1 - p = 0.95

so the answer must be (n pick k) * p^1 * q^(n-k).

If you forget about the above, then you can perhaps build the answer from the basics. In a sequence of 4 men, what is the probability of getting: SFFF where S is success and F is failure. The answer is (0.05)*(0.95)^3. This case meets the criterion i.e. one success in four but there are three other possibilities as well:

FSFF, FFSF, FFFS.

So the total probability over all the cases is 4 * (0.05) * (0.95)^3. This happens to be the same quantity as reported by the binomial distribution formula above.