Sheldon ross -Conditional probability

Question

Comments

what is the correct answer given in the ross?

---

0.093 is the answer.

Answer 1

let 52 cards are divided into 2 halves name A and B

therefore A consists 26 cards and B consists 26 cards

given that an ace card is drawn from A ===> B might have 3 or 2 or 1 or 0 Ace Cards in it's 26 cards.

now ace card from A is combined with B ==> B might have 4 or 3 or 2 or 1 Ace Cards in it's 27 cards

let E_i = B have i no.of ACE cards and E = drawing ace card from B

what is the probability of Ace card from B = ( P( E₁ ) * P( E | E₁ ) ) +  ( P( E₂ ) * P( E | E₂ ) ) +  ( P( E₃ ) * P( E | E₃ ) ) +  ( P( E₄ ) * P( E | E₄ ) ) 

             = ($\frac{1}{4} * \frac{1}{27}$) + ($\frac{1}{4} * \frac{2}{27}$) + ($\frac{1}{4} * \frac{3}{27}$) + ($\frac{1}{4} * \frac{4}{27}$)

            =  ( $\frac{1}{4} $ ) * ( $\frac{1}{27}$ + $\frac{2}{27}$ + $\frac{3}{27}$ + $\frac{4}{27}$)

            = ( $\frac{1}{4} $ ) * ( $\frac{10}{27}$ ) 

            = ( $\frac{5}{54} $ ) 

Comments

Can you Please tell me how p(E1)=P(E2)=P(E3)=1/4 ??

---

let take 2 coins and toss them at a time

Total possible outcomes = { TT,TH,HT,HH }

no.of T's = either 2 or 1 or 0

what is the probabilities of no.of T's equal to 2 =  $ \frac{1}{4} $ ( favorable = TT )

what is the probabilities of no.of T's equal to 1 =  $ \frac{2}{4} $ ( favorable = TH,HT )

 what is the probabilities of no.of T's equal to 0 =  $ \frac{1}{4} $ ( favorable = HH ) 

---

no.of Ace's = either 4 or 3 or 2 or 1

what is the probabilities of no.of Ace's equal to 4 =  $ \frac{1}{4} $  ( favorable = 4 Ace's )

what is the probabilities of no.of Ace's equal to 3 =  $ \frac{1}{4} $  ( favorable = 3 Ace's ) 

what is the probabilities of no.of Ace's equal to 2 =  $ \frac{1}{4} $  ( favorable = 2 Ace's )

what is the probabilities of no.of Ace's equal to 1 =  $ \frac{1}{4} $  ( favorable = 1 Ace's )

---

Can you Please tell me how p(E1)=P(E2)=P(E3)=1/4 ??

So, what is P(Ei)? it is that set B has i number of ACEs.
how many such cases possible set-B
having only 1 ACE
                  2 ACEs
                  3 ACEs
                  4 ACEs

So, there are four cases and one the case will definitely occur and all cases have an equal probability of occurrence i.e 1/4, which is used in the answer.

Answer 2

There are total $4$ aces. Since there was an ace in first deck, that means the number of aces in second deck possibly are : $0,1,2,3$.

Let $P_{i}$ be the probability that the second deck contains $i(0,1,2,3)$ aces. Let $E$ be the event that you draw an ace from the second deck.

So basically we have to calculate $P(E|P_{i})(P_{i})$

$P_{0}$ means that we have second deck where no ace is present initially, this means $P_{0} = \binom{48}{26}/\sum_{i=0}^{3}N_{i}$

Where $N_{i}$ is the total number of half decks with i aces in it. Similarly we can calculate other $P_{i}$'s. Now  $P(E|P_{i}) = (i+1)/27$

Just put in the values and calculate.

Comments

No need to be only one ace in first half(it should be at least one ace).