In this experiment with cards numbered 1,3,5,7 two cards are selected without re

Question

In this experiment with cards numbered 1,3,5,7 two cards are selected without replacement. The number of times that a sum has occurred is recorded. Think of taking out one card, setting it down and selecting a second card without replacing the first card.

A) what is the probability of a sum of a six according to the experiment?

B) What is the experimental probability of at least a sum of 8 occuring?

C) What is the theoretical P(sum of 6 or 8)

D) How does the theoretical probability of a sum of a 6 compare to the experimental probability of a sum of a 6? Why did this happen?

Sum 4 6 8 10 12 n 8 9 6 2 3

Explanation / Answer

Total nuber of experiment :

8+9+6+2+3 = 28

A)

Out of 28 experiment, 9 have sum 6 so the probability of a sum of a six according to the experiment is

P(sum 6) = 9/28 = 0.3214

B)

Out of 28 experiment, 6+2+3 = 11 experiment has sum at least 8 so

P(sum at least 8) = 11/28 = 0.3929

C)

Following table shows all possible samples and corresponding probabilites:

Following table shows the pdf of S:

So

P(sum of 6 or 8) = 2/12 + 4/12 = 6/12 = 0.5

D)

Theoritical probability:

P(sum of 6) = 2/12 = 1/6 = 0.1667

Samples Sum, S P(S=s) 1 3 4 1/12 1 5 6 1/12 1 7 8 1/12 3 1 4 1/12 3 5 8 1/12 3 7 10 1/12 5 1 6 1/12 5 3 8 1/12 5 7 12 1/12 7 1 8 1/12 7 3 10 1/12 7 5 12 1/12

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