# Thread: probability 52 card problem

1. ## probability 52 card problem

A standard deck of 52 cards is shuffled, and each of 3 people are asked to select a card at
random from the deck. Each time a card is selected, it is returned to the deck and reshuffled.
What is the probability that
(a) all 3 people select different cards?
(b) exactly one of the 3 people selects a heart?
(c) the first person selects a heart or the last person selects an ace?

heres what im tinking... let me know if this seens logical
a) (1/52)(1/51)(1/50)

b)(1/52) (1/52) (1/4)

c)

2. Originally Posted by zackgilbey
A standard deck of 52 cards is shuffled, and each of 3 people are asked to select a card at random from the deck. Each time a card is selected, it is returned to the deck and reshuffled. What is the probability that
(a) all 3 people select different cards?
(b) exactly one of the 3 people selects a heart?
(c) the first person selects a heart or the last person selects an ace?
Because the card is returned to the deck after each draw, the events are independent.
This means for c) $P\left( {H_1 \cup A_3 } \right) = P\left( {H_1 } \right) + P\left( {A_3 } \right) - P\left( {H_1 \cap A_3 } \right) = \left( {\frac{{13}}{{52}}} \right) + \left( {\frac{4}{{52}}} \right) - \left( {\frac{{13}}{{52}}} \right)\left( {\frac{4}{{52}}} \right)$

3. Hello, zackgilbey!

A standard deck of 52 cards is shuffled,
and each of 3 people are asked to select a card at random from the deck.
Each time a card is selected, it is returned to the deck and reshuffled.

What is the probability that
(a) all 3 people select different cards?

The first person can select any card: . $\frac{52}{52}$

The second must select one of the other 51 cards: . $\frac{51}{52}$

The third must select one of the other 50 cards: . $\frac{50}{52}$

$P(\text{3 different cards}) \;=\;\frac{52}{52}\cdot\frac{51}{52}\cdot\frac{50} {52} \;=\;\frac{1275}{1352}$

(b) exactly one of the 3 people selects a heart?

There are 3 choices for the person who selects the Heart.
He gets a Heart with probability: . $\frac{13}{52} \,=\,\frac{1}{4}$

The other two people must select two non-hearts: . $\frac{39}{52}\cdot\frac{39}{52} \,=\,\frac{9}{16}$

$P(\text{exactly one Heart}) \;=\;3\cdot\frac{1}{4}\cdot\frac{9}{16} \;=\;\frac{27}{64}$

(c) the first person selects a heart or the last person selects an ace?

The first selects a Heart: . $\frac{13}{52}$

The second can select any card . . . We can ignore him!

The third selects an Ace: . $\frac{4}{52}$

We also have: . $P(\text{1st Heart }and\text{ 3rd Ace}) \;=\;\frac{13}{52}\cdot\frac{4}{52} \;=\;\frac{1}{52}$

Formula: . $P(A \vee B) \:=\:P(A) + P(B) - P(A \wedge B)$

We have: . $P(\text{1st Heart }\vee\text{ 3rd Ace}) \;=\;P(\text{1st Heart}) + P(\text{3rd Ace}) - P(\text{1st Heart }\wedge\text{ 3rd Ace})$

Therefore: . $P(\text{1st Heart }\vee\text{ 3rd Ace}) \;=\;\frac{13}{52} + \frac{4}{52} - \frac{1}{52} \;=\;\frac{16}{52} \;=\;\frac{4}{13}$