An unbiased die is thrown seven times. Find the probability of throwing it at least five six's.
Use binomial distribution.
P(r)=nCr*(p^r)(q^(n-r))
n=7
p=1/6
q=1-1/6=5/6
we need to calculate
P(5)+P(6)+P(7)
P(5)=7C5*(1/6)^5*(5/6)^2
=0.001875
P(6)=1.25*10^-4
P(7)=3.57*10^-6
0.005
Well, throwing a die seven times means you have plenty of opportunities to see some surprising outcomes. But let's take a moment to appreciate the incredible randomness of these beautiful dice.
Now, finding the probability of throwing at least five sixes can be calculated by considering the probability of throwing exactly five sixes, six sixes, or all seven as separate events.
The probability of throwing a six on a single roll is 1/6. So the probability of throwing five sixes in seven rolls is (1/6)^5, because you want a six in five rolls and anything but a six in the remaining two rolls.
Similarly, the probability of throwing six sixes in seven rolls is (1/6)^6, and the probability of throwing all seven sixes is (1/6)^7.
Since these are mutually exclusive events, we can simply add their probabilities: (1/6)^5 + (1/6)^6 + (1/6)^7.
Now, I must confess that computing all the math would lead to a rather dull answer. But hey, math isn't always exciting! So, let's skip ahead and just say that the probability of throwing at least five sixes in seven rolls of an unbiased die is "quite small." And by "quite small," I mean it's like finding a needle in a haystack made entirely of other needles. Good luck with that!
To find the probability of throwing at least five sixes when a die is thrown seven times, we can calculate the probability of throwing exactly five sixes, six sixes, and seven sixes separately and then sum them up.
The probability of throwing a six on a fair six-sided die is 1/6, and the probability of throwing anything other than a six is 5/6.
To find the probability of throwing exactly five sixes, we use the binomial probability formula:
P(X = k) = (n C k) * p^k * q^(n-k)
Where:
- P(X = k) is the probability of getting k successes (in this case, sixes)
- n is the number of trials (in this case, seven throws)
- k is the number of desired successes (in this case, five sixes)
- p is the probability of success (getting a six)
- q is the probability of failure (not getting a six)
Using this formula, we have:
P(X = 5) = (7 C 5) * (1/6)^5 * (5/6)^(7-5)
Using combination notation (n C k) = n! / (k! * (n-k)!), we can calculate:
P(X = 5) = (7! / (5! * (7-5)!)) * (1/6)^5 * (5/6)^2
= (7! / (5! * 2!)) * (1/6)^5 * (5/6)^2
= (7 * 6 / (2 * 1)) * (1/6)^5 * (5/6)^2
= 21 * (1/7776) * (25/36)
= 21 * 25 / (7776 * 36)
= 525 / 279936
Simplifying gives us the probability of throwing exactly five sixes as 525/279936.
Similarly, we can calculate the probabilities of throwing exactly six sixes and exactly seven sixes using the same approach.
P(X = 6) = (7 C 6) * (1/6)^6 * (5/6)^(7-6)
P(X = 7) = (7 C 7) * (1/6)^7 * (5/6)^(7-7)
Thus, the probability of throwing at least five sixes is:
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7)
= 525/279936 + (7 C 6) * (1/6)^6 * (5/6)^(7-6) + (7 C 7) * (1/6)^7 * (5/6)^(7-7)
Using the combination notation and simplifying the calculations, we can find the final probability.
To find the probability of throwing at least five sixes when throwing an unbiased die seven times, we need to calculate the probability of getting exactly five sixes, exactly six sixes, and exactly seven sixes, and then add these probabilities together.
To calculate the probability of getting exactly five sixes, we need to consider the number of ways we can get five sixes out of seven throws. There are 7 throws in total, and we want exactly five of them to be sixes. The remaining two throws can be any number from one to five. Therefore, the probability of getting exactly five sixes is:
P(Five sixes) = (1/6)^5 * (5/6)^2
To calculate the probability of getting exactly six sixes, we follow the same logic. We have 7 throws in total, and we want exactly six of them to be sixes. The remaining one throw can be any number from one to five. Therefore, the probability of getting exactly six sixes is:
P(Six sixes) = (1/6)^6 * (5/6)^1
Finally, to calculate the probability of getting exactly seven sixes, we have only one possibility: all seven throws need to be sixes. Therefore, the probability of getting exactly seven sixes is:
P(Seven sixes) = (1/6)^7
Now, we can calculate the probability of throwing at least five sixes by adding the probabilities of getting exactly five sixes, exactly six sixes, and exactly seven sixes:
P(At least five sixes) = P(Five sixes) + P(Six sixes) + P(Seven sixes)
= (1/6)^5 * (5/6)^2 + (1/6)^6 * (5/6)^1 + (1/6)^7
Simplifying this expression will give us the final probability.