Question

Paper is often tested for "burst strength" "teart strength" say we classify these strengths as low, middle,and high. Then ,after exammining 100 piecesof paper, we find the folloing

Burst Strength
_________________
Tear A 1 A2 A3
strength (low) (middle) (high)
_______________________________________________
B1(low) 7 11 13
B2(middle) 11 21 9
B3(high) 12 9 7
_____________________________________________
If we select one of the pieces random. what are the probability that it has the following charactcristics:
(a) A1
(b) A3 intersection B2
(c) A2uinion B3
(d)A1, given that it is B2
(e) B1,given that it is A3

Answer 1

The probabilities can be calculated based on the data provided. (a) Probability of having A1 is 0.31. (b) Probability of having A3 intersection B2 is 0.09. (c) Probability of having A2 union B3 is 0.3. (d) Probability of having A1 given that it is B2 is 0.5238. (e) Probability of having B1 given that it is A3 is 1.7143.

Step-by-step explanation:

The probabilities can be calculated based on the data provided.

(a) To find the probability of having A1, we need to sum up the values in the first row of the table: 7 + 11 + 13 = 31. The total number of pieces of paper is 100. So the probability of having A1 is 31/100 = 0.31 (or 31%).

(b) To find the probability of having A3 intersection B2, we need to find the value in the table where A3 and B2 intersect. From the table, this value is 9. So the probability is 9/100 = 0.09 (or 9%).

(c) To find the probability of having A2 union B3, we need to sum up the values in the second row and the third column: 21 + 9 = 30. The total number of pieces of paper is 100. So the probability is 30/100 = 0.3 (or 30%).

(d) To find the probability of having A1 given that it is B2, we need to consider only the values in the second column (B2) and find the value in the first row (A1). From the table, this value is 11. So the probability is 11/21 = 0.5238 (or 52.38%).

(e) To find the probability of having B1 given that it is A3, we need to consider only the values in the third row (A3) and find the value in the first column (B1). From the table, this value is 12. So the probability is 12/7 = 1.7143 (or 171.43%).