To calculate the probability that exactly 3 out of 7 randomly selected bacteria cells are capable of cellular replication, you can use the binomial probability formula:
P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)
Where:
- n is the total number of trials (in this case, 7 bacteria cells selected).
- k is the number of successful trials (in this case, 3 bacteria capable of replication).
- p is the probability of a single trial being a success (in this case, the probability that a randomly selected cell is capable of replication).
First, calculate the probability of success (p). Out of the 36 bacteria cells, 12 are not capable of replication, so there are 36 - 12 = 24 bacteria capable of replication. Therefore, p = 24/36 = 2/3.
Now, plug these values into the formula:
P(X = 3) = (7 choose 3) * (2/3)^3 * (1 - 2/3)^(7 - 3)
To calculate "7 choose 3," you can use the binomial coefficient formula:
(7 choose 3) = 7! / (3! * (7 - 3)!) = (7 * 6 * 5) / (3 * 2 * 1) = 35
Now, plug all the values into the formula:
P(X = 3) = 35 * (2/3)^3 * (1/3)^4
Calculate each part:
P(X = 3) = 35 * (8/27) * (1/81)
Now, multiply these values:
P(X = 3) = (35 * 8 * 1) / (27 * 81) = 280 / 2187 ≈ 0.1281
So, the probability that exactly 3 out of 7 randomly selected bacteria cells are capable of cellular replication is approximately 0.1281, or about 12.81%.