A calculator is programmed to generate a random integer between 1 and 10. In six trials, what is the probability that the number 3 is generated exactly twice?
this is binomial ... three (t) or not three (n)
p(t) = .1 ... p(n) = .9
(n + t)^6 = n^6 + 6 n^5 t + 15 n^4 t^2 + ... + 6 n t^5 + t^6
evaluate the 3rd term for the solution ... 15 * .9^4 * .1^2 = ?
There are only 8 numbers "between" 1 and 10
so the prob of getting a 3 = 1/8
prob of not getting a 3 = 7/8
prob(event as stated) = C(6,2) (1/8)^2 (7/8)^4
= 15(1/64)(2401/4096) = ...
To determine the probability of a specific outcome occurring exactly a certain number of times in a series of trials, we need to use the binomial probability formula.
The formula for the binomial probability is:
P(x) = (nCx) * p^x * q^(n-x)
Where:
- P(x) is the probability of getting exactly x successes
- n is the number of trials
- x is the number of successful outcomes we're interested in
- p is the probability of success on a single trial
- q is the probability of failure on a single trial (q = 1 - p)
- nCx is the binomial coefficient which represents the number of ways to choose x items from a set of n items (nCx = n! / (x! * (n-x)!)
In this case, n = 6 (number of trials), x = 2 (number of times the number 3 should be generated), p = 1/10 (probability of generating the number 3 in a single trial), and q = 1 - p = 9/10.
Plugging in these values, we can calculate the probability as follows:
P(2) = (6C2) * (1/10)^2 * (9/10)^(6-2)
Calculating each part:
6C2 = 6! / (2! * (6-2)!) = 6! / (2! * 4!) = (6 * 5 * 4!) / (2! * 4!) = (6 * 5) / (2 * 1) = 15
(1/10)^2 = 1/100
(9/10)^(6-2) = (9/10)^4 = 6561/10,000
Plugging these values back into the formula:
P(2) = 15 * (1/100) * (6561/10,000) = 15 * 6561 / (100 * 10,000) = 98415 / 1,000,000 = 0.098415
Therefore, the probability that the number 3 is generated exactly twice in six trials is approximately 0.098415 or 9.84%.
To find the probability of generating the number 3 exactly twice in six trials, we need to calculate two things:
1) the number of ways we can get exactly two 3's out of six trials, and
2) the probability of getting a 3 in each trial.
1) The number of ways to get exactly two 3's out of six trials can be calculated using combinations. We can use the formula for combinations, which is given by:
C(n, k) = n! / (k!(n-k)!)
In this case, we want to calculate C(6, 2) since we want to choose 2 trials out of 6 where the number 3 is generated. Plugging the numbers into the formula, we get:
C(6, 2) = 6! / (2!(6-2)!) = 6! / (2!4!) = (6 * 5 * 4!) / (2!4!) = (6 * 5) / 2 = 15
So, there are 15 ways to get exactly two 3's out of six trials.
2) The probability of getting a 3 in each trial is 1/10, since the calculator is programmed to generate a random integer between 1 and 10.
Now, to find the probability of getting exactly two 3's out of six trials, we need to multiply the number of ways to get exactly two 3's (15) by the probability of getting a 3 in each trial (1/10), and then raise it to the power of 2 (since we want exactly two 3's). Finally, we multiply this by the probability of not getting a 3 in the remaining trials (9/10) raised to the power of the remaining trials (6 - 2 = 4).
Therefore, the probability of getting exactly two 3's in six trials is:
(15 * (1/10)^2 * (9/10)^4) = (15/100) * (81/10000) = 1215/100000
Hence, the probability of generating the number 3 exactly twice in six trials is 1215/100000.