During a lottery, several lots are to be selected. A is the event that

Question

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During a lottery, several lots are to be selected. A is the event that a certain subset of these lots is selected, and B is the event that another subset of these lots is selected, such that:

• the probability of event A occurring is 2/5.
• the probability of event B occurring is 4/5.
• the probability of the union of events A and B occurring is 1.
• the intersection of events A and B is an event with four desirable outcomes.

In the following table, identify the total number of lots and the number of desirable outcomes in event B. Make only one selection in each column.

Sajjad1994 · Accepted Answer

Official Explanation

The probability of the union of events A and B occurring equals the probability of event A occurring, plus the probability of event B occurring, minus the probability of the intersection of events A and B occurring. You know the value of all of these probabilities except the last one. The probability of the intersection of events A and B equals the number of desirable outcomes in the intersection of A and B, which is 4, over the number of total outcomes in the lottery—let’s call that t. Thus, you have:

P (AUB) = P (A) + P (B) - P (A∩B)

1=\frac{2}{5}+\frac{4}{5}-\frac{4}{t}

1=\frac{2t-4t-20}{5t}

5t=6t-20

t=20

So the correct answer in the first column = the total number of lots is 20.

Next, let the number of desirable outcomes in event B be b. In that case, you have:

P(B) =\frac{b}{20}

\frac{4}{5}=\frac{b}{20}

4=\frac{b}{4}

b=16

So the correct answer in the second column is: the number of desirable outcomes in event B is 16.

AdelaRachel · Answer

hi Sajjad1994, could you help to explain this question?

Yoga_Venkatesh · Answer

Can someone explain why P(A) and P(B) are considered as 2/5 and 4/5 while the question mentioned as 1/3 and 1/3?

Gemmie · Answer

Event A               X Event A          Subtotal

Event B           1/5                        3/5                4/5

X Event B        1/5                         0                  1/5

Subtotal          2/5                         3/5                 1

4 desirable outcomes for the intersection of event A & B
=>  4 = \frac{1}{5}  * total slots
=> total slots = 20

Desirable outcomes for B:  \frac{4}{5} * 20 = 16

Purnank · Answer

given numbers are so good just observe them, for which of those following you will ge P(B) as 4/5
16 and 20 those are your answers.

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Number of lots	Desirable outcomes in B
		15
		16
		17
		18
		19
		20

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