Joe has a very large pile of batteries. 60% of the batteries are Energizer brand

Question

Joe has a very large pile of batteries. 60% of the batteries are Energizer brand and 40% are Duracell brand. Of the Energizer brand, 50% are AA. 30% are AAA, and 20% are D batteries. Of the Duracell brand, 70% are AA, 15% are AAA, and 15% are D batteries. For both brands, 1% of the A A batteries are dead, 2% of the AAA batteries are dead, and 3% of the D batteries are dead. Joe randomly selects a battery from the pile and it turns out to be dead. What is the probability that the dead battery is an Energizer A A battery?

Explanation / Answer

Back-up Theory

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then

Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(1)

P(B) = {P(B/A) x P(A)} + {P(B/AC) x P(AC)}………………………………….(2)

If A is made up of k mutually and collectively exhaustive sub-events, A1, A2,.....Ak,

P(B) = sum over i = 1 to k of {P(B/Ai) x P(Ai)} ………………………………..(3)

P(A/B) = P(B/A) x { P(A)/P(B)}……………………………..………………….(4)

Now, to work out solution,

60% of Joe’s pile of batteries are Energizer brand (E) => probability that a randomly picked battery from the pile is Energizer brand is 0.6.

Thus, we have, P(E) = 0.6 and P(D) = 0.4, where D represents Duracell brand.

Similarly, P(E is A) = 0.5, P(E is AA) = 0.3, P(E is D) = 0.2,

P(D is A) = 0.7, P(D is AA) = 0.15 and P(D is D) = 0.15.

Hence, probability that a randomly picked battery from the pile is Energizer brand A = 0.6 x 0.5 = 0.3 and so on.

Now, let the following events be defined as follows:

A1 – Energiser A, A2 – Energiser AA, A3 – Energiser D,

A4 – Duracell A, A5 – Duracell AA, A6 – Duracell D and

B – battery is dead.

Then we have

P(A1) = 0.3, P(A2) = 0.18, P(A3) = 0.12, P(A4) = 0.28, P(A5) = 0.06, P(A6) = 0.06, and

[Note that these probabilities add to 1. This is good check to detect gross errors]

P(B/A1) = 0.01, P(B/A2) = 0.02, P(B/A3) = 0.03,

P(B/A4) = 0.01, P(B/A5) = 0.02, P(B/A6) = 0.03.

And hence, [vide (3) under Back-up Theory],

P(B) = 0.003 + 0.0036 + 0.0036 + 0.0028 + 0.0012 + 0.0018 = 0.0160

We want probability the battery is EAA given that it is dead.

i.e., P(A2/B) = P(B/A2)P(A2)/P(B) [vide (3) under Back-up Theory]

= (0.02 x 0.18)/0.016 = 0.225 ANSWER

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

---

---

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