Answer:
[tex]P(X_s^c|X_F) =0.2[/tex]
[tex]P(X_s^c|X_F) =0.31[/tex]
[tex]P(X_s^c|X_F) =0.331[/tex]
Step-by-step explanation:
From the given information:
Let represent [tex]X_F[/tex] as the first player getting an ace
Let [tex]X_S[/tex] to be the second player getting an ace and
[tex]\sim X_S[/tex] as the second player not getting an ace.
So;
The probabiility of the second player not getting an ace and the first player getting an ace can be computed as;
[tex]P(\sim X_S| X_F) = 1 - P(X_S|X_F)[/tex]
[tex]P(X_S|X_F) = \dfrac{P(X_SX_F)}{P(X_F)}[/tex]
Let's determine the probability of getting an ace in the first player
i.e
[tex]P(X_F) = 1 - P(X_F^c)[/tex]
[tex]= 1 -\dfrac{(^{2n-2}_n)}{(^{2n}_n)}}[/tex]
[tex]= 1 - \dfrac{n-1}{2(2n-1)}[/tex]
[tex]= \dfrac{3n-1}{4n-2} --- (1)[/tex]
To determine the probability of the second player getting an ace and the first player getting an ace.
[tex]P(X_sX_F) = \text{ (distribute aces to both ) and (select the left over n-1 cards from 2n-2 cards}[/tex][tex]P(X_sX_F) = \dfrac{2(^{2n-2}C_{n-1})}{^{2n}C_n}[/tex]
[tex]P(X_sX_F) = \dfrac{n}{2n -1}---(2)[/tex]
[tex]P(X_s|X_F) = \dfrac{2}{1}[/tex]
[tex]P(X_s|X_F) = \dfrac{2n}{3n -1}[/tex]
Thus, the conditional probability that the second player has no aces, provided that the first player declares affirmative is:
[tex]P(X_s^c|X_F) = 1- \dfrac{2n}{3n -1}[/tex]
[tex]P(X_s^c|X_F) = \dfrac{n-1}{3n -1}[/tex]
Therefore;
for n= 2
[tex]P(X_s^c|X_F) = \dfrac{2-1}{3(2) -1}[/tex]
[tex]P(X_s^c|X_F) = \dfrac{1}{6 -1}[/tex]
[tex]P(X_s^c|X_F) = \dfrac{1}{5}[/tex]
[tex]P(X_s^c|X_F) =0.2[/tex]
for n= 10
[tex]P(X_s^c|X_F) = \dfrac{10-1}{3(10) -1}[/tex]
[tex]P(X_s^c|X_F) = \dfrac{9}{30 -1}[/tex]
[tex]P(X_s^c|X_F) = \dfrac{9}{29}[/tex]
[tex]P(X_s^c|X_F) =0.31[/tex]
for n = 100
[tex]P(X_s^c|X_F) = \dfrac{100-1}{3(100) -1}[/tex]
[tex]P(X_s^c|X_F) = \dfrac{99}{300 -1}[/tex]
[tex]P(X_s^c|X_F) = \dfrac{99}{299}[/tex]
[tex]P(X_s^c|X_F) =0.331[/tex]
