In a raffle there are four winning tickets from 12, What is the probability from six tickets there are two winners?
Thank you in advance.
The number of getting two winning tickets out of six is 6! / (2! x 4!) = 6x5/2 = 15.
Consider the probability of getting one of those 15 outcomes - say just the first two are winners, and the final four are losers. The chances of that happening - remember you're choosing WITHOUT replacement, so the odds of picking a winning ticket change every time - are (4/12) x (3/11) x (8/10) x (7/9) x (6/8) x (5/7) = (1/33).
Now consider the probability of getting a different combination of winners and losers in which there are still just two winners - say the 2nd and 4th one are the winners. The odds of THAT happening are (8/12) x (4/11) x (7/10) x (3/9) x (6/8) x (5/7). The denominators are exactly the same as before, and the numerators have just all been shuffled around - so the product of them all is the same as before, namely (1/33).
Try any other combination, and you'll see that exactly the same thing happens again. So the answer should be the total number of ways of getting two winners out of six times the probability of any one of them happening, namely 15 x (1/33). So I reckon the answer is (15/33).
To find the probability of having two winners out of six tickets in a raffle with four winning tickets out of 12 total tickets, we can use the binomial probability formula.
The binomial probability formula is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of getting exactly k successes,
C(n, k) is the number of combinations of n items taken k at a time,
p is the probability of success on a single trial,
n is the number of trials (in this case, the number of tickets drawn), and
k is the number of successful outcomes we are interested in (in this case, the number of winning tickets).
In this case, we have n = 6 (six tickets drawn), k = 2 (two winning tickets), and p = 4/12 (probability of drawing a winning ticket).
Let's calculate the probability step-by-step:
1. Calculate the number of combinations of 6 tickets taken 2 at a time:
C(6, 2) = 6! / (2! * (6 - 2)!) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15
2. Calculate the probability of drawing a winning ticket:
p = 4/12 = 1/3
3. Substitute the values into the binomial probability formula:
P(X = 2) = C(6, 2) * (1/3)^2 * (1 - 1/3)^(6 - 2)
P(X = 2) = 15 * (1/3)^2 * (2/3)^4
4. Calculate the final probability:
P(X = 2) = 15 * (1/9) * (16/81)
P(X = 2) = 240/729
Therefore, the probability of having two winners out of six tickets in the given raffle is approximately 0.3289, or 32.89%.
To find the probability of getting two winners out of six tickets in a raffle with four winning tickets out of twelve, we need to calculate the combination.
The total number of ways to select 6 tickets out of 12 is given by the combination formula:
nCr = n! / (r!(n-r)!)
Where n is the total number of tickets and r is the number of tickets you want to choose.
In this case, we want to find the number of ways to choose 2 winning tickets out of 4 and 4 losing tickets out of 8 (12-4=8).
So, the number of ways to choose 2 winning tickets out of 4 is given by the combination formula:
4C2 = 4! / (2!(4-2)!) = 6
And the number of ways to choose 4 losing tickets out of 8 is:
8C4 = 8! / (4!(8-4)!) = 70
Now, we need to calculate the probability, which is the number of favorable outcomes (getting 2 winners out of 6 tickets) divided by the total number of outcomes (choosing any 6 tickets out of 12):
Probability = Number of ways to choose 2 winners out of 4 * Number of ways to choose 4 losing tickets out of 8 / Total number of ways to choose 6 tickets out of 12
Probability = (4C2 * 8C4) / (12C6) = (6 * 70) / 924 = 420 / 924 ≈ 0.4545
So, the probability of getting two winners out of six tickets in this raffle is approximately 0.4545 or 45.45%.