1.if 3 out of 24 marked scripts have mathematical error and 5 are to be randomly selected for audit,what is the probability that none of the marked scripts with mathematical errors will be chosen?

2.ten identical PCs are in the inventory of a dealer,one has a hidden defect.if 3 are to be shipped,and the computers are selected in such a way that each has the same probability of being shipped,find
I.the number of ways in which 3 of the 10 PCs can be selected for shipment.
II.the total number of ways in which 3 of the 10PCs can be selected so that the one with the hidden defect will be included for shipment.
III.the probability that the computer with the hidden defect will be shipped.

1. Number of marked scripts with mathematical errors=3.

2. Number of marked scripts without mathematical errors= 24-3 = 21
3. The probability of selecting 5 scripts without mathematical errors = 5/21 = 0.24.

To answer the first question, we need to calculate the probability that none of the marked scripts with mathematical errors will be chosen when 5 scripts are randomly selected for audit.

Step 1: Calculate the probability of selecting a script without a mathematical error.
Out of the 24 scripts, 3 have mathematical errors, so there are 24 - 3 = 21 scripts without mathematical errors.
The probability of selecting a script without a mathematical error is 21/24.

Step 2: Calculate the probability of selecting 5 scripts without any mathematical errors.
Since each selection is independent, we can multiply the probabilities together.
The probability of selecting 5 scripts without any mathematical errors is (21/24) * (21/24) * (21/24) * (21/24) * (21/24).

Therefore, the probability that none of the marked scripts with mathematical errors will be chosen is (21/24) * (21/24) * (21/24) * (21/24) * (21/24).

To answer the second question, we need to calculate the number of ways in which 3 of the 10 PCs can be selected for shipment, as well as the number of ways in which the one with the hidden defect will be included.

I. The number of ways in which 3 of the 10 PCs can be selected for shipment is given by the combination formula. Using the formula:

Number of ways = nCr = n! / (r!(n-r)!)

In this case, n = 10 (total number of PCs) and r = 3 (number of PCs to be selected for shipment). Plugging in the values:

Number of ways = 10! / (3!(10-3)!) = 10! / (3!7!)

Simplifying this expression will give you the total number of ways.

II. To calculate the total number of ways in which 3 of the 10 PCs can be selected so that the one with the hidden defect will be included, you can consider the one with the hidden defect as already chosen. Therefore, you need to select 2 PCs from the remaining 9. Using the combination formula again:

Number of ways = 9! / (2!(9-2)!).

III. Finally, to find the probability that the computer with the hidden defect will be shipped, you divide the number of ways in which it is included (calculated in part II) by the total number of ways (calculated in part I).

Probability = (Number of ways with hidden defect included) / (Total number of ways)

Hope this helps! Let me know if you have any other questions.