0
$\begingroup$

What is the probability that after pulling out of a card deck 3 heart cards, that the 4th card will be also a heart? There are 52 cards in the deck and there is no replacement.

$$P(4\text{th heart} | 3 \text{ first hearts}) = \frac{P(4\text{th heart}\cap 3\text{ first hearts})}{P(3\text{ firsthearts})}$$

How do you calculate this probability: $P(4\text{th heart}\cap 3\text{ first hearts})$

$\endgroup$
2
  • $\begingroup$ How many cards are in the deck? How are you drawing them? With replacement or not? $\endgroup$
    – 5xum
    Commented Sep 3, 2015 at 10:10
  • $\begingroup$ 52 cards no replacement $\endgroup$
    – DoubleOseven
    Commented Sep 3, 2015 at 10:11

4 Answers 4

Reset to default
4
$\begingroup$

You are overcomplicating things. You can use the more intuitive meaning of conditional probability

The probability of $P(X|Y)$ is the probability that $X$ will happen if you know that $Y$ already happened.

Well, if you know that the first three cards were hearts, then the probability that the fourth card will be a heart is equal to the probability of drawing a heart out of a deck of $49$ cards of which $10$ are hearts.

Of course, you can also go with the original attempt. Then the event

The first three cards were hearts and the fourth card was a heart

Is the same event as

The first four cards were all hearts.

$\endgroup$
5
  • $\begingroup$ Thank you that makes perfect sense! :) $\endgroup$
    – DoubleOseven
    Commented Sep 3, 2015 at 10:15
  • $\begingroup$ In your first explanation the prob is 10/49, but in the second explanation the prob is (13*12*11*10)/(52*51*50*49). Am I thinking wrong? $\endgroup$
    – DoubleOseven
    Commented Sep 3, 2015 at 10:23
  • $\begingroup$ @Mathguy007 You are correct. However, the number $\frac{10}{49}$ is the probability $P(X|Y)$, and the second number is the probability $P(X\cap Y)$ $\endgroup$
    – 5xum
    Commented Sep 3, 2015 at 10:25
  • $\begingroup$ arghhh Thank you for your help :) $\endgroup$
    – DoubleOseven
    Commented Sep 3, 2015 at 10:28
  • $\begingroup$ @Mathguy007 If you divide the second number by $P(Y) = \frac{13\cdot 12\cdot 11}{52\cdot 51\cdot 50}$, you get the first number. $\endgroup$
    – 5xum
    Commented Sep 3, 2015 at 10:29
0
$\begingroup$

Initially you have 13 hearts in 52 cards deck so probability of first heart is 13/52.

Then you have 12 hearts in 51 cards deck => probability of second heart given first heart is 12/51. And probability of first two hearts is $\frac{13}{52}\cdot\frac{12}{51}$.

Then you can proceed up to 4 pulled cards.

$\endgroup$
0
$\begingroup$

Hint:

After pulling out $3$ hearts there are $49$ cards left and $10$ of them are hearts (preassumed that it is a normal deck). So the probability that at that stage a heart will be drawn is...

$\endgroup$
0
$\begingroup$

In the beginning, you have a $\frac {13}{52}$ chance of choosing a Heart. After that is removed, you now have $12$ cards out of $51$. So you now have a $\frac {12}{51}$ chance of choosing a Heart.

Continuing this on, we get $$\frac {13}{52}\times\frac {12}{51}\times\frac {11}{50}\times\frac {10}{49}=\boxed{\frac {11}{4165}}$$

Which comes out to be about $0.3\%$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .