Please answer CLEARLY and LEGIBLE .4.10. In an office there are two boxes of thu

Question

Please answer CLEARLY and LEGIBLE .4.10. In an office there are two boxes of thumb drives: Box A1 contains seven 100 ves and three 500 GB drives, and box A2 contains two 100 GB drives and eight 500 GB drives. A person is handed a box at random with prior probabilities P(A1)-| and P(A2-, possibly due to the boxes, respective locations. A drive is then selected at random and the event B occurs if it is a 500 GB drive. Using an equally likely assumption for each drive in the selected box, compute P(A1|B) and P(A2|B)

Explanation / Answer

Using conditional probability formula we have:

P(A1 | B) = P(A1 and B)/P(B)

Using total probability law, we have:

P(B) = P(B and A1) + P(B and A2)

Also,

P(B | A1) = P(A1 and B)/P(A1)

Since box A1 has 7 100gb drives and 3 500gb drives, so

P(B | A1) = 3/(7+3) = 0.3

So,

0.3 = P(A1 and B)/P(A1)

Also it is given that P(A1) = 2/3

So,

P(A1 and B) = 0.3*P(A1) = 0.3/(2/3) = 0.45

Similarly,

P(B | A2) = P(A2 and B)/P(A2)

Since box A2 has 2 100gb drives and 8 500gb drives, so

P(B | A2) = 8/(2+8) = 0.8

So,

0.8 = P(A2 and B)/P(A2)

Also it is given that P(A2) = 1/3

So,

P(A2 and B) = 0.8*P(A2) = 0.8*(1/3) = 0.267

So,

P(B) = 0.45 + 0.267 = 0.717

So, finally we have:

P(A1 | B) = P(A1 and B)/P(B) = 0.45/0.717 = 0.627

P(A2 | B) = P(A2 and B)/P(B) = 0.267/0.717 = 0.373

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.