probability 2 cards question

R

Redmoon

New member
Joined
Mar 23, 2022
Messages
15
Question
Here are five number cards. 3 6 7 2 5 Two of the five cards are picked at random. Work out the probability that the product of the two cards will be greater than 20.

Hey So what I did was create a table which adds each possible combination of the two cards up

so like this
3 6 7 2 5
3. blank 9 10 5 8
6. 9 13 8 11
7. 10 13. 9 12
2. 5. 8. 10. 7
5. 8. 11. 12 7


So total out is 20

over 20. =. 0/16 divided by the 2 cards = 0/18 as the final answer?

I think its really the last bit of my answer I think might be wrong can any one help?
 
Redmoon said:
Question
Here are five number cards. 3 6 7 2 5 Two of the five cards are picked at random. Work out the probability that the product of the two cards will be greater than 20.

Hey So what I did was create a table which adds each possible combination of the two cards up

so like this
3 6 7 2 5
3. blank 9 10 5 8
6. 9 13 8 11
7. 10 13. 9 12
2. 5. 8. 10. 7
5. 8. 11. 12 7


So total out is 20

over 20. =. 0/16 divided by the 2 cards = 0/18 as the final answer?

I think its really the last bit of my answer I think might be wrong can any one help?
Click to expand...
Why do you tabulate the sums when the problem asks about the products?

The last bit of the answer does not look right to me either -- can you explain your approach there?
 
Please note that you can use the "insert table" icon to create a readable table.
 
This problem is easy if you stop to think BEFORE you do any computation

Is multiplication dependent on the order of the factors? No.

If we pick two number cards, how many distinct factors do we have? Two.

How many ways can we pick two distinct items from five distinct items without regard to order?

[math]\dbinom{5}{2} = \dfrac{5!}{2! * (5 - 2)!} = \dfrac{5 * 4}{2} = 10.[/math]
Now you could build a table of the ten possible products, but there is a short cut.

There are four products that have 2 as the smallest factor. The largest such product is [imath]2 \times 7 = 14.[/imath] So none of those products > 20.

There are three products that have 3 as the smallest factor. The largest such product is 21. So how many of these products is larger than 20?

The smallest remaining product is [imath]5 \times 6[/imath]. How many of those products are larger than 20?

So the probability of a product > 20 is what?

Combinatorics is more about careful thinking than about computation. The table method is valid, but it may require more work and thus more opportunity for error.
 
blamocur said:
Why do you tabulate the sums when the problem asks about the products?
Click to expand...
Thanks so much for pointing this out as I never notice the obvious. Someone tells me multiplication, the last thing I think of is addition.
 
Redmoon said:
Question
Here are five number cards. 3 6 7 2 5 Two of the five cards are picked at random. Work out the probability that the product of the two cards will be greater than 20.
Click to expand...
@Redmoon, There are only ten pair to consider.
Multiply the two numbers in each pair together.
How many of those products are greater than twenty?
Post your answer.
 
pka said:
@Redmoon, There are only ten pair to consider.
Multiply the two numbers in each pair together.
How many of those products are greater than twenty?
Post your answer.
Click to expand...
Thanks for you help, what I got was 4 pairs are greater than 20.

7 and five = 35
6 and 5 = 30
6 and 7 = 42
3 and 7 = 21

Then to get the final answer I divided the 4 pairs I got over the ten possible out comes so the final answer I got is 2/5 would this be correct or am I still going wrong somewhere thanks?
 
Redmoon said:
Thanks for you help, what I got was 4 pairs are greater than 20.

7 and five = 35
6 and 5 = 30
6 and 7 = 42
3 and 7 = 21

Then to get the final answer I divided the 4 pairs I got over the ten possible out comes so the final answer I got is 2/5 would this be correct or am I still going wrong somewhere thanks?
Click to expand...
Four out of ten is [imath]\dfrac{2}{5}[/imath] and is correct.
 
You must log in or register to reply here.
Top