Answer:
0.001145
Step-by-step explanation:
Given that you are certain to get 3 jacks when selecting 51 cards from a shuffled deck.
When we draw 3 cards from 51 cards, we keep one card aside
The card can be either Jack or non Jack
Prob (the left over card to be Jack) = P(A) = [tex]\frac{4}{52}[/tex]
Prob (the left over card to be non jack ) = P(B) = [tex]\frac{48}{52}[/tex]
A and B are mutually exclusive and exhaustive
Let drawing 3 Jacks be the event C
P(C) = P(AC)+P(BC)
[tex]P(AC) = P(A)*P(C/A)\\= \frac{4}{52} (3/51C3)\\=1.108(10^{-5}[/tex]
P(BC) = P(B)*P(C/B)\\
= \frac{28}{52} (4/51C3)\\
=1.034(10^{-4}
Adding we get
P(C)
= [tex](1.108+10.343)(10^{-5} \\=0.001145[/tex]
