(a) A container has 16 smart phones numbered 1 to 16. Except for the labels, all

Question

(a) A container has 16 smart phones numbered 1 to 16. Except for the labels, all the devices are identical. A device is drawn at random from the container. Let A, B, C and D represent the events of picking a device with number in the sets IA, IB, Ic, and ID, respectively. Are A, B, C and D independent?

IA = {1, 2, 3, 4, 5, 6, 7, 8}

IB = {1, 2, 3, 4, 9, 10, 11, 12}

IC = {1, 5, 6, 7, 9, 10, 11, 13}

ID = {1, 2, 7, 8, 10, 11, 15, 16}

(b) Suppose there are three bits transmitted, which are known to be all 1s or all 0s. In addition, suppose that the first and second bits are known to be a 1s with probability 1/3. Given this information, what is the probability that the third bit is 1?

Explanation / Answer

(a)

P(A) = P(B) = P(C) = P(D) = 8/16 = 1/2

P(A and B) = P(A and C) = P(A and D) = P(B and C) = P(B and D) = P(C and D) = 4/16 = 1/4

Thus,

P(A and B) = P(A) P(B) = 1/4

P(A and C) = P(A) P(C) = 1/4

P(A and D) = P(A) P(D) = 1/4

P(B and C) = P(B) P(C) = 1/4

P(B and D) = P(B) P(D) = 1/4

P(C and D) = P(C) P(D) = 1/4

Hence,

the events A, B, C and D are indepndent

(b)

There are 3 transmitted bits, which are known to be all 1s or all 0s.

Suppose that the first bits and second bits are known to be 1s with probability 1/3

What is the probability that the third bit is 1?

Probability that third bit is 1

= Probability that first bit is 1 * Probability that second bit is 1

= 1/3 * 1/3

= 1/9

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