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 5 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?

Explanation / Answer

total events =8+9+6+2+3=28

a)  probability of a sum of a six according to the experiment =9/28 (as 9 events when 6 appears out of 28)

b)xperimental probability of at least a sum of 8 occuring =(6+2+3)/28 =11/28

c)  theoretical P(sum of 5 or 8) =P(sum of 5)+P(Sum of 8) =P(no possible comination for 5)+P(first card 1 and second seven+first card 7 and second seven+first card 3 and second 5+first card 5 and second 3)

=0+(1/4)*(1/3)+1/4)*(1/3)+1/4)*(1/3)+1/4)*(1/3) =4/12 =1/3

D) theoritical  probability of a sum of a 6 =P(first card 1 and second 5+first card 5 and second1)

=(1/4)*(1/3)+1/4)*(1/3) =1/6

while those with experiment is 9/28

this happens because of randomness of the process which will mitigate the differnce and experimental probability will be nearly same to theoritical probability once sample size grew larger,.

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