# Maths help

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

can someone help me with this question pls?

There are 9 counters in a bag.

There is an even number on 3 of the counters. There is an odd number on 6 of the counters

Three counters are going to be taken at random from the bag.

The numbers on the counters will be added together to give the total

Find the probability that the total is an odd number

There are 9 counters in a bag.

There is an even number on 3 of the counters. There is an odd number on 6 of the counters

Three counters are going to be taken at random from the bag.

The numbers on the counters will be added together to give the total

Find the probability that the total is an odd number

0

reply

Report

#2

three numbers can add to a total of an odd number if:

- two numbers are even and one is odd (e.g. 2 + 2 + 3 = 7)

- all three numbers are odd (e.g. 3 + 3 + 3 = 9)

to find the probability of getting a total of an odd number, you would need to find the total probability of both events above.

you can assume that the three counters taken are not put back in the bag. this would mean that, if you were to take 3 odd numbers out, there would be 9 in the bag on the first draw, 8 on the second, and so on.

for 3 odd numbers:

on the first draw, there are 6 odd numbers and 3 even numbers, so the probability of getting an odd number is 6/9.

on the second, there are 5 odd and 3 even, so the probability of getting an odd number is 5/8.

on the third, there are 4 odd and 3 even, so the probability of getting an odd number is 4/7.

the probability of getting 3 odd numbers is the probability of all three events above happening at the same time. to find the probability of multiple events at the same time, you can multiply their probabilities.

(6/9 * 5/8 * 4/7 = 120/504 = 5/21)

for 2 even numbers and 1 odd number:

this can happen in 3 ways - 1 even, 1 odd, 1 even; 1 odd, 1 even, 1 even; 1 even, 1 even, 1 odd.

even on first draw: there are 3 evens and 6 odds, so the probability is 3/9.

odd on second draw: there are 2 evens and 6 odds, so the probability is 6/8.

even on third draw: there are 2 evens and 5 odds, so the probability is 2/7.

the probability of all three happening is the product of each probability.

(3/9 * 6/8 * 2/7 = 1/14)

the probability for the other combinations is the same as the one above.

the second 6/9 * 3/8 * 2/7, while the third gives 3/9 * 2/8 * 6/7.

the numbers in the top and bottom of each fraction are simply rearranged, and so the product is the same.

the probability of getting an odd number totaled from 2 evens and 1 odd, in either one of these 3 ways, is found by adding the probabilities of each combination.

(1/14 + 1/14 + 1/14 = 3/14)

the probability of getting an odd number totaled from 2 evens and 1 odd, or from getting 3 odd numbers, is found by adding the probabilities of both.

(3/14 + 5/21 = 9/42 + 10/42 = 19/42)

- two numbers are even and one is odd (e.g. 2 + 2 + 3 = 7)

- all three numbers are odd (e.g. 3 + 3 + 3 = 9)

to find the probability of getting a total of an odd number, you would need to find the total probability of both events above.

you can assume that the three counters taken are not put back in the bag. this would mean that, if you were to take 3 odd numbers out, there would be 9 in the bag on the first draw, 8 on the second, and so on.

for 3 odd numbers:

on the first draw, there are 6 odd numbers and 3 even numbers, so the probability of getting an odd number is 6/9.

on the second, there are 5 odd and 3 even, so the probability of getting an odd number is 5/8.

on the third, there are 4 odd and 3 even, so the probability of getting an odd number is 4/7.

the probability of getting 3 odd numbers is the probability of all three events above happening at the same time. to find the probability of multiple events at the same time, you can multiply their probabilities.

(6/9 * 5/8 * 4/7 = 120/504 = 5/21)

for 2 even numbers and 1 odd number:

this can happen in 3 ways - 1 even, 1 odd, 1 even; 1 odd, 1 even, 1 even; 1 even, 1 even, 1 odd.

even on first draw: there are 3 evens and 6 odds, so the probability is 3/9.

odd on second draw: there are 2 evens and 6 odds, so the probability is 6/8.

even on third draw: there are 2 evens and 5 odds, so the probability is 2/7.

the probability of all three happening is the product of each probability.

(3/9 * 6/8 * 2/7 = 1/14)

the probability for the other combinations is the same as the one above.

the second 6/9 * 3/8 * 2/7, while the third gives 3/9 * 2/8 * 6/7.

the numbers in the top and bottom of each fraction are simply rearranged, and so the product is the same.

the probability of getting an odd number totaled from 2 evens and 1 odd, in either one of these 3 ways, is found by adding the probabilities of each combination.

(1/14 + 1/14 + 1/14 = 3/14)

the probability of getting an odd number totaled from 2 evens and 1 odd, or from getting 3 odd numbers, is found by adding the probabilities of both.

(3/14 + 5/21 = 9/42 + 10/42 = 19/42)

2

reply

(Original post by

three numbers can add to a total of an odd number if:

- two numbers are even and one is odd (e.g. 2 + 2 + 3 = 7)

- all three numbers are odd (e.g. 3 + 3 + 3 = 9)

to find the probability of getting a total of an odd number, you would need to find the total probability of both events above.

you can assume that the three counters taken are not put back in the bag. this would mean that, if you were to take 3 odd numbers out, there would be 9 in the bag on the first draw, 8 on the second, and so on.

for 3 odd numbers:

on the first draw, there are 6 odd numbers and 3 even numbers, so the probability of getting an odd number is 6/9.

on the second, there are 5 odd and 3 even, so the probability of getting an odd number is 5/8.

on the third, there are 4 odd and 3 even, so the probability of getting an odd number is 4/7.

the probability of getting 3 odd numbers is the probability of all three events above happening at the same time. to find the probability of multiple events at the same time, you can multiply their probabilities.

(6/9 * 5/8 * 4/7 = 120/504 = 5/21)

for 2 even numbers and 1 odd number:

this can happen in 3 ways - 1 even, 1 odd, 1 even; 1 odd, 1 even, 1 even; 1 even, 1 even, 1 odd.

even on first draw: there are 3 evens and 6 odds, so the probability is 3/9.

odd on second draw: there are 2 evens and 6 odds, so the probability is 6/8.

even on third draw: there are 2 evens and 5 odds, so the probability is 2/7.

the probability of all three happening is the product of each probability.

(3/9 * 6/8 * 2/7 = 1/14)

the probability for the other combinations is the same as the one above.

the second 6/9 * 3/8 * 2/7, while the third gives 3/9 * 2/8 * 6/7.

the numbers in the top and bottom of each fraction are simply rearranged, and so the product is the same.

the probability of getting an odd number totaled from 2 evens and 1 odd, in either one of these 3 ways, is found by adding the probabilities of each combination.

(1/14 + 1/14 + 1/14 = 3/14)

the probability of getting an odd number totaled from 2 evens and 1 odd, or from getting 3 odd numbers, is found by adding the probabilities of both.

(3/14 + 5/21 = 9/42 + 10/42 = 19/42)

**euphrosynay**)three numbers can add to a total of an odd number if:

- two numbers are even and one is odd (e.g. 2 + 2 + 3 = 7)

- all three numbers are odd (e.g. 3 + 3 + 3 = 9)

to find the probability of getting a total of an odd number, you would need to find the total probability of both events above.

you can assume that the three counters taken are not put back in the bag. this would mean that, if you were to take 3 odd numbers out, there would be 9 in the bag on the first draw, 8 on the second, and so on.

for 3 odd numbers:

on the first draw, there are 6 odd numbers and 3 even numbers, so the probability of getting an odd number is 6/9.

on the second, there are 5 odd and 3 even, so the probability of getting an odd number is 5/8.

on the third, there are 4 odd and 3 even, so the probability of getting an odd number is 4/7.

the probability of getting 3 odd numbers is the probability of all three events above happening at the same time. to find the probability of multiple events at the same time, you can multiply their probabilities.

(6/9 * 5/8 * 4/7 = 120/504 = 5/21)

for 2 even numbers and 1 odd number:

this can happen in 3 ways - 1 even, 1 odd, 1 even; 1 odd, 1 even, 1 even; 1 even, 1 even, 1 odd.

even on first draw: there are 3 evens and 6 odds, so the probability is 3/9.

odd on second draw: there are 2 evens and 6 odds, so the probability is 6/8.

even on third draw: there are 2 evens and 5 odds, so the probability is 2/7.

the probability of all three happening is the product of each probability.

(3/9 * 6/8 * 2/7 = 1/14)

the probability for the other combinations is the same as the one above.

the second 6/9 * 3/8 * 2/7, while the third gives 3/9 * 2/8 * 6/7.

the numbers in the top and bottom of each fraction are simply rearranged, and so the product is the same.

the probability of getting an odd number totaled from 2 evens and 1 odd, in either one of these 3 ways, is found by adding the probabilities of each combination.

(1/14 + 1/14 + 1/14 = 3/14)

the probability of getting an odd number totaled from 2 evens and 1 odd, or from getting 3 odd numbers, is found by adding the probabilities of both.

(3/14 + 5/21 = 9/42 + 10/42 = 19/42)

damn this is a long , lowkey complicated but u managed to explain it clearly . Thanks.

0

reply

Report

#4

(Original post by

damn this is a long , lowkey complicated but u managed to explain it clearly . Thanks.

**jadeb456**)damn this is a long , lowkey complicated but u managed to explain it clearly . Thanks.

0

reply

(Original post by

it is indeed long.. but i hope it helped to understand the processes in solving the problem

**euphrosynay**)it is indeed long.. but i hope it helped to understand the processes in solving the problem

0

reply

Report

#6

I set this up using a 3 stage tree diagram, one stage for each counter. P (even) = 3/9, P (odd) = (6/9) for first selection. I assumed non replacement.Need either 1 or 3 odd counters for an odd result.So P(e,o,e OR e,e,o OR o,e,e OR o,o,o) = 29/42I’d love to know if anyone else agrees?!

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top