Three players are each dealt, in a random manner, five cards from a deck containing 52 cards. four of the 52 cards are aces. find the probability that at least one person receives exactly two aces in their five cards.

The probability that at least one person receives exactly two aces in their five cards is 0.1194.

Step-by-step explanation:

We define the following events:

A: person A receives exactly two aces in their five cards.

B: person B receives exactly two aces in their five cards.

C: person C receives exactly two aces in their five cards.

We need to calculate P(A∪B∪C). According to the principle of inclusion and exclusion:

P(A∪B∪C)= P(A)+P(B)+P(C)-P(A∩B)-P(A∩C)-P(B∩C)+P(A∩B∩C)

Total outcomes is equal to because there 52 cards and we select 5 of them.

The probability of each event is given by:

P(A)=P(B)=P(C)===0.0399

where the possible outcomes are 2 aces and 3 of the other cards.

Then, the intersections of the events mean that one player gets 2 aces and the other gets 2 aces too.

P(A∩B)=P(B∩C)=P(A∩C)==0.0001

Finnaly,

P(A∩B∩C)=0 because it is not possible that the three events occur at the same time due to there are only 4 aces.

 P(A∪B∪C)= 0.0399+0.0399+0.0399-0.0001-0.0001-0.0001+0

 P(A∪B∪C)=0.1194

The probability that at least one person receives exactly two aces in their five cards is 0.1194.