probability problem about cards

Oct 2008
897
3
Bristol, England
Hi, this is my question:
Sean has a set of cards numbered 1,2,3 and 4. He picks a card at random and does not replace it. He then picks another card at random from the remaining cards.

A) What is the probability that neither of the two cards that Sean picked was 4?

im not sure but would i do, 3/4 x 2/3= 6/12 = 3/6 = 1/2?
Answer being 1/2?

Thanks!
 
Dec 2009
3,120
1,342
Hi, this is my question:
Sean has a set of cards numbered 1,2,3 and 4. He picks a card at random and does not replace it. He then picks another card at random from the remaining cards.

A) What is the probability that neither of the two cards that Sean picked was 4?

im not sure but would i do, 3/4 x 2/3= 6/12 = 3/6 = 1/2?
Answer being 1/2?

Thanks!
Yes, that's the quick way..

\(\displaystyle \binom{4}{2}=6\) is the number of ways to choose 2 cards from 4.

If the pair does not contain 4, there are \(\displaystyle \binom{3}{2}=3\) pairs of two cards not containing 4.

Listing the options... 1,2....1,3....2,3 do not contain 4

1,4....2,4,....3,4 contain 4

half of the two-card pairings do not contain 4.
 
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Oct 2008
897
3
Bristol, England
Thanks alot!!

the next question says:

B) What is the probability that the numbers on the two cards that Sean picked add up to 4?

im not sure how to do this, would it be:
3/4 x 2/3 = 1/2 again??
 
Dec 2009
3,120
1,342
Thanks alot!!

the next question says:

B) What is the probability that the numbers on the two cards that Sean picked add up to 4?

im not sure how to do this, would it be:
3/4 x 2/3 = 1/2 again??
In order to add to 4, the cards must be 1 and 3.
That's just 1 out of the 6 selections of 2 cards.

If you count arrangements, there are two arrangements.....1,3 and 3,1

but there are are 4(3)=12 arrangements.

To do it by multiplying probabilities,
there is a 1/4 probability the first is 1 and a 1/3 probability the second is 3.
Similarly there is a 1/4 probability the first is 3 and a 1/3 probability the second is a 1.

That's \(\displaystyle \frac{1}{4}\ \frac{1}{3}+\frac{1}{4}\ \frac{1}{3}\)

Or, there's a 2 in 4 chance the first is a 1 or 3 and a 1 in 3 chance the second is the other

\(\displaystyle P=\frac{2}{4}\ \frac{1}{3}\)
 
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Oct 2008
897
3
Bristol, England
In order to add to 4, the cards must be 1 and 3.
That's just 1 out of the 6 selections of 2 cards.

If you count arrangements, there are two arrangements.....1,3 and 3,1

but there are are 4(3)=12 arrangements.

To do it by multiplying probabilities,
there is a 1/4 probability the first is 1 and a 1/3 probability the second is 3.
Similarly there is a 1/4 probability the first is 3 and a 1/3 probability the second is a 1.

That's \(\displaystyle \frac{1}{4}\ \frac{1}{3}+\frac{1}{4}\ \frac{1}{3}\)

Or, there's a 2 in 4 chance the first is a 1 or 3 and a 1 in 3 chance the second is the other

\(\displaystyle P=\frac{2}{4}\ \frac{1}{3}\)

oh, so i could just write ' the probability the two cards sean picks up add up to 4 is 2/4'?
 
Dec 2009
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1,342
oh, so i could just write ' the probability the two cards sean picks up add up to 4 is 2/4'?
No, 2/4 is the probability that the first number is one of the numbers that will give a sum of 4.

You need both numbers.

Suppose the first number was 3.
Then the probability the 2nd one is 1 is 1/3.

Multiply the probabilities of the first being 3 and the 2nd being 1

this is \(\displaystyle \frac{1}{4}\ \frac{1}{3}=\frac{1}{12}\)

there is also a 1/12 probability of getting a 1 followed by a 3.

Summing the probabilities gives 2/12=1/6.

This is the probability of getting the sequence 1,3 or 3,1
which are the only ways to get a sum of 4 from the two cards.

There are other ways to do it, but the probability will be 1/6

If you list the possibilities

12 13 14
21 23 24
31 32 34
41 42 43

Only 2 of these 12 arrangements give a sum of 4

12 13 14
23 24
34

Only 1 of these 6 selections gives a sum of 4
 
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Oct 2008
897
3
Bristol, England
oh i understand now, thanks!

the last part is:

Sean now picks a third card from the remaining two cards.

C) Calculate the probability that the numbers on the first two cards add up to the number on the third card.

im not sure what to do on this one aswell, so could i have some more help please?
 
Dec 2009
3,120
1,342
oh i understand now, thanks!

the last part is:

Sean now picks a third card from the remaining two cards.

C) Calculate the probability that the numbers on the first two cards add up to the number on the third card.

im not sure what to do on this one aswell, so could i have some more help please?
Ok,

first examine what the possibilities are...

the first two cards can be

1 and 2
1 and 3
1 and 4
2 and 3
2 and 4
3 and 4

This rules out four of these pairings since the highest number card is 4

The first two cards can only be 1 and 2 or 1 and 3.

If the first two cards are 1 and 2 then the 3rd card must be 3.
If the first two cards are 1 and 3 then the third card must be 4.

Now that the problem is clear, you can use arrangements or selections to calculate the probabilities.

1,2,3 has a \(\displaystyle \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}\) probability

because there is a 1 in 4 chance of picking 1 first,
a 1 in 3 chance of picking 2 second
and a 1 in 2 chance of picking 4 third.

Also, the 2 could come before 1, but the probability is the same.

So the probability of a 1 and 2 followed by 3 is \(\displaystyle 2\frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}=\frac{1}{12}\)

Working out the probability of 1,3,4 or 3,1,4 is the same procedure.

Adding up the probabilities, you get 1/12+1/12=2/12=1/6

You can list the options also

1,2,3
1,2,4
1,3,2
1,3,4
1,4,2
1,4,3
2,1,3
2,1,4
2,3,1
2,3,4
2,4,1
2,4,3
3,1,2
3,1,4
3,2,1
3,2,4
3,4,1
3,4,2
4,1,2
4,1,3
4,2,1
4,2,3
4,3,1
4,3,2

There are 24 arrangements and 4 of these are in the required order
 
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Oct 2008
897
3
Bristol, England
Ok,

first examine what the possibilities are...

the first two cards can be

1 and 2
1 and 3
1 and 4
2 and 3
2 and 4
3 and 4

This rules out four of these pairings since the highest number card is 4

The first two cards can only be 1 and 2 or 1 and 3.

If the first two cards are 1 and 2 then the 3rd card must be 3.
If the first two cards are 1 and 3 then the third card must be 4.

Now that the problem is clear, you can use arrangements or selections to calculate the probabilities.

1,2,3 has a \(\displaystyle \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}\) probability

because there is a 1 in 4 chance of picking 1 first,
a 1 in 3 chance of picking 2 second
and a 1 in 2 chance of picking 4 third.

Also, the 2 could come before 1, but the probability is the same.

So the probability of a 1 and 2 followed by 3 is \(\displaystyle 2\frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}=\frac{1}{12}\)

Working out the probability of 1,3,4 or 3,1,4 is the same procedure.

Adding up the probabilities, you get 1/12+1/12=2/12=1/6

You can list the options also

1,2,3
1,2,4
1,3,2
1,3,4
1,4,2
1,4,3
2,1,3
2,1,4
2,3,1
2,3,4
2,4,1
2,4,3
3,1,2
3,1,4
3,2,1
3,2,4
3,4,1
3,4,2
4,1,2
4,1,3
4,2,1
4,2,3
4,3,1
4,3,2

There are 24 arrangements and 4 of these are in the required order

when adding 1/4 + 1/3 +1/2 how does it = 1/12??
wouldn't it be 1/9?
 
Dec 2009
3,120
1,342
when adding 1/4 + 1/3 +1/2 how does it = 1/12??
wouldn't it be 1/9?
We sum the probabilities of each way to get the desired result.

Each way involves the multiplication of the probabilities at each stage of choosing cards for that way.

In other words......

1,2,3 is 1 followed by 2 followed by 3.

This is a "successful" sequence from the perspective of the number of the 3rd card
being the sum of the numbers of the first two cards.

This is.... 1 and 2 and 3 together.

So, to calculate the probability of this happening, we calculate

(i) the probability that 1 came first, this is 1/4
(ii) the probability that 2 came second, this is 1/3 as there are three cards left and 2 is one of them
(iii) the probability that 3 came third.

The probability that 1,2,3 happened in that order is the product of those "stage-by-stage" probabilities.

The probability of 1 and 2 and 3 in that order is \(\displaystyle \frac{1}{4}\ \frac{1}{3}\ \frac{1}{2}\)

That's where we multiply probabilities.

Now we can also have 2,1,3 so that probability must be worked out.
Also 1,3,4 and 3,1,4 are successful outcomes.

Hence we calculate the probabilities for those 4 ways of getting a successful
outcome and then add them all up

In other words....1,2,3 or 2,1,3 or 1,3,4 or 3,1,4

is a successful outcome.

Or.... add probabilities
And...multiply probabilities

is a way to think of it.

You can check from the sequences listed how the probabilities should work out
 
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