A box contains four numbered balls (1, 2, 2, and 3). We will randomly select two
ID: 3065022 • Letter: A
Question
A box contains four numbered balls (1, 2, 2, and 3). We will randomly select two balls from the box (without replacement). (Hint: since the selections are without replacement, this means, for example, that if the ball labelled "1" is selected on the first selection, it can not be selected again on the second selection.) Question 1 (1 point) The outcome of interest is the number on each of the two balls we select. How many outcomes are contained in the sample space? (A sample space consists of only unique outcomes. It does not consider how frequently each outcome occurs. Therefore, no outcome should be listed more than once in the sample space.) Your Answer: Question 2 (1 point) Let X be the sum of the numbers on the two selected balls. Find the probability that x-3. Report your answer to 2 decimal places. (Hint: first consider which of the outcomes in the previous question satisfy X 3.,) Your Answer: Save Question 3 (1 point) Let X be the sum of the numbers on the two selected balls. What is the probability that the first selection is not 1 if we know that x-47 Report your answer to 2 decimal places.Explanation / Answer
The sample space consists of {(1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2)}
X denotes the sum of the two outcomes. Let X1 be the outcome of the first draw and X2 be the outcome of the second draw.
P (X=3) = P(X1=1) x P(X2=2) + P(X1=2) x P(X2=1)
= (1/4) * (2/3) + (2/4)*(1/3) = 2/6=0.33
P(X=4) = P(X1=1) x P(X2=3) + P(X1=2) x P(X2=2)
+ P(X1=3) x P(X2=1) = (1/4)*(1/3) + (2/4)*(1/3) + (1/4)*(1/3) = 4/12 = 1/3
Out-of these possibilities, probability that X1 is not equal to 1 = P(X1=2) x P(X2=2) + P(X1=3) x P(X2=3) = (2/4)*(1/3) + (1/4)*(1/3) = 3/12 = 1/4
So the probability that the first selection is not 1, given that X=4 is (1/4) / (1/3) = (3/4) = 0.75
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