the daily production of two factories A and B are 300 and 200 bottle respectively .2% of the product o f each factory are found to be defective .if 6 defective bottle are randomly selected from a daily production of A and B .find the probability that 4 of the defective bottles are from A
I don't get it
To find the probability that 4 of the defective bottles are from factory A, we need to calculate the probability of selecting 4 defective bottles from factory A and 2 defective bottles from factory B.
Let's break down the problem step by step:
Step 1: Calculate the total number of defective bottles from factory A and factory B.
- The daily production of factory A is 300 bottles, and 2% of them are defective, so the number of defective bottles from factory A is 300 * 2% = 6.
- Similarly, the daily production of factory B is 200 bottles, and 2% of them are defective, so the number of defective bottles from factory B is 200 * 2% = 4.
Step 2: Calculate the probability of selecting 4 defective bottles from factory A and 2 defective bottles from factory B.
- To do this, we need to calculate the probability of selecting 4 defective bottles from factory A, which can be done using the combination formula.
- The combination formula is given by: C(n, r) = n! / (r! * (n - r)!)
- In this case, n is the total number of defective bottles from factory A (6) and r is the number of defective bottles we want to select from factory A (4).
So, the probability of selecting 4 defective bottles from factory A is:
P(4 defective bottles from A) = C(6, 4) / (C(10, 6) * C(10, 2))
Using the combination formula, C(6, 4) = 6! / (4! * (6 - 4)!) = 15
C(10, 6) = 10! / (6! * (10 - 6)!) = 210
C(10, 2) = 10! / (2! * (10 - 2)!) = 45
So, the probability is:
P(4 defective bottles from A) = 15 / (210 * 45) = 15 / 945 ≈ 0.0159
Therefore, the probability that 4 of the defective bottles are from factory A is approximately 0.0159, or 1.59%.