Probability question on sets: What's the probability of randomly choosing an even number from the set B′=B∪{b1,b2}?

[BogdanSzymanski]

Let A and B be the following sets: [imath]A=\{1,2,3,4,5\}[/imath], [imath]B=\{2,4,6,7,8\}[/imath]. Let [imath]b_1[/imath] and [imath]b_2[/imath] be two randomly selected numbers from the set B. What's the probability of randomly choosing an even number from the set [imath]B'=B\cup \{b_1,b_2\}[/imath]?

 

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[imath]\;[/imath]

 

[Dr.Peterson]

Let A and B be the following sets: [imath]A=\{1,2,3,4,5\}[/imath], [imath]B=\{2,4,6,7,8\}[/imath]. Let [imath]b_1[/imath] and [imath]b_2[/imath] be two randomly selected numbers from the set B. What's the probability of randomly choosing an even number from the set [imath]B'=B\cup \{b_1,b_2\}[/imath]?

Please check that you copied this correctly. As written, set A doesn't appear to be involved in the question; and [imath]B'=B\cup \{b_1,b_2\}=B[/imath].

 

[Steven G]

I 2nd that B′=B∪{b₁,b₂}=B since b₁and b₂ are in B.

For example C = {1,2,3,1,3,4} = {1,2,3,4}. Repeating an element in a set does not change the set nor the cardinality of the set.

I am assuming that set A was needed for another part of this problem(?)

 

[BogdanSzymanski]

I apologise, it should have been [imath]A'=A\cup \{b_1, b_2\}[/imath]

 

[Dr.Peterson]

I apologise, it should have been [imath]A'=A\cup \{b_1, b_2\}[/imath]

Now please read post #2 and do what it says: Show us your own work, so we can see what help you need (assuming that reading the problem correctly doesn't fix everything).

 

[ImMAD]

Lets start by considering what happens when we add just b_1 to A. Clearly, 2/5 of the time it is in A, 2/5 of the time it is not in A and is even, and 1/5 of the time it is not in A and is odd. Then, obtaining an even number in the above cases for A ∪ b_1 is 2/5, 1/2, and 1/3 respectively, and the total probability of obtaining an even number for A ∪ b_1 will be 2/5 * 2/5 + 2/5 * 1/2 + 1/5 * 1/3 = 32/75.

Now, when we want to add a second number b_2 to A, we have to consider 3 cases.
Case 1, b_1 was already in A, which will occur 2/5 of the time.
Case 2, b_1 was not in A, and was even, which will occur 2/5 of the time.
Case 3, b_1 was not in A, and was odd, which will occur 1/5 of the time.

Case 1 means we can just apply the result of adding b_1 to A again (this time under the label b_2), so the probability of obtaining a even number out of A' will be 32/75.
Case 2 means A ∪ b_1 = {1,2,3,4,5,6} or {1,2,3,4,5,8}, resulting in a 3/5 chance of b_2 being already in the set, 1/5 chance that b_2 is not in the set, and even, and 1/5 chance that b_2 is not in the set, and odd. Then, obtaining an even number in the above case will be 1/2, 4/7, and 4/7 respectively.
Case 3 means A ∪ b_1 = {1,2,3,4,5,7}, resulting in a 3/5 chance of b_2 being already in the set, and a 2/5 chance that b_2 is not in the set, and even. Then, obtaining an even number in the above case will be 1/3, and 3/7 respectively.

Adding all this information together (try finishing this calculation yourself), we obtain a probability of obtaining a even number from A' equal to 1198/2625, approximately equal to 46%.