Answer:
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 [tex]C^{52}_{5}[/tex] because there 52 cards and we select 5 of them.
The probability of each event is given by:
P(A)=P(B)=P(C)=[tex]\frac{C^{4}_{2}.C^{48}_{3}}{C^{52}_{5}}[/tex]=[tex]\frac{\frac{4!}{2!2!} .\frac{48!}{3!45!}}{\frac{52!}{5!47!}}[/tex]=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)=[tex]\frac{C^{4}_{2}.C^{2}_{2}.C^{48}_{1}}{C^{52}_{5}}[/tex]=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.
