Question

Slips of paper with the numbers 2, 4, 6, and 8 are placed into a hat. A slip is drawn, recorded,
and then returned to the hat. Then, a second slip is drawn and recorded. Use this description
to find:
i) The probability that the sum of the two drawn slips is at least 13
ii) The probability that the slips are the same number
iii) Whether the events described in (i) and (ii) are mutually exclusive. (Explain)
iv) The probability of the sum of two drawn slips being even OR less than nine.

Answer 1

Sample space :-

(2,2) (4,4) (6,6) (8,8) (2,4) (2,6) (2,8) (4,6) (4,8) (6,8) (4,2) (6,2) (8,2) (6,4) (8,4) (8,6) = 16 samples

(1) sum of the two drawn slips is atleast 13 = 13 and 13 above

Samples whose sum is at least 13 or above = (6,8) (8,6) 8,8) = 3 samples

Probability that the sum of samples being drawn is 13 and above = 3/16

(2) samples with same numbers = (2,2) (4,4) (6,6) (8,8) = 4 samples

Probability that the slips are same number = 4/16 = 1/4

(3) No, the events 1 and 2 are not mutually exclisive because exclusive events cannot occur at the same time but event 1 and event 2 can occur at the same time, as the sum of the two slips at least 13 is also occurring in the same numbers event, i.e., (8,8) is common in both events.

(4) as we know that sum of two even numbers is always even.

And OR means, either this or that.

So, we have to choose the sample in which the sum of two numbers is even or less than 9 = all the samples = 16 (because all the samples has even sum which includes sum of numbers less than 9 also)

Probability that the sum of the numbers less than 9 or being even = 16/16 = 1

NOTE :- If the last part would have asked that find the probability of the sum of the two drawn slips being even AND less than 9

Then the samples would have been = (2,2) (4,4) (2,4) (2,6) (4,2) (6,2) = 6 samples

Then the probability would have been = 6/16 = 3/8  

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