Paul has 8 cards. [GCSE MATHS PAST PAPER QUESTION]

I see the community here is cool so i thought i'd ask a question:

Paul has 8 cards.
There is a number on each card.

2,3,3,4,5,5,5,5

Paul takes at random 3 cards.
He adds together the 3 numbers on the card to get a total T

Work out the probability that T is an odd number.

Thanks in advance.

Reply 1

anujsr

3

Original post by Kasperov

I see the community here is cool so i thought i'd ask a question:

Paul has 8 cards.
There is a number on each card.

2,3,3,4,5,5,5,5

Paul takes at random 3 cards.
He adds together the 3 numbers on the card to get a total T

Work out the probability that T is an odd number.

Thanks in advance.

There are two cards with even numbers and six cards with odd numbers. For the sum to be odd Paul must have two even and one odd card at any time or have all three cards that have odd numbers. This could happen in the following combinations:

Odd even even
even odd even
even even odd
odd odd odd

The probability of each combination is shown below:

odd even even = (6/8) x (2/7) x (1/6) = 1/28
even odd even = (2/8) x (6/7) x (1/6) = 1/28
even even odd = (2/8) x (1/7) x (6/6) = 1/28
odd odd odd = (6/8) x (5/7) x (4/6) =5/14
now add all of these probabilities together : 1/28 +1/28 + 1/28 + 5/14 =13/28

Hope that helps

Reply 2

nadiakms

14

odd,odd,even
even,odd,odd
odd,even,odd

1- p(even possibilities) = odd values
1- 3(6/8 x 5/7 x 2/6)
1-3(60/336)
1-(180/336) = 156/336 = 13/28

Reply 3

Kasperov

OP

Original post by anujsr

There are two cards with even numbers and six cards with odd numbers. For the sum to be odd Paul must have two even and one odd card at any time or have all three cards that have odd numbers. This could happen in the following combinations:

Odd even even
even odd even
even even odd
odd odd odd

The probability of each combination is shown below:

odd even even = (6/8) x (2/7) x (1/6) = 1/28
even odd even = (2/8) x (6/7) x (1/6) = 1/28
even even odd = (2/8) x (1/7) x (6/6) = 1/28
odd odd odd = (6/8) x (5/7) x (4/6) =5/14
now add all of these probabilities together : 1/28 +1/28 + 1/28 + 5/14 =13/28

Hope that helps

Original post by nadiakms

odd,odd,even
even,odd,odd
odd,even,odd

1- p(even possibilities) = odd values
1- 3(6/8 x 5/7 x 2/6)
1-3(60/336)
1-(180/336) = 156/336 = 13/28

Thanks guys

Reply 4

rocteacher

1

Original post by Kasperov

I see the community here is cool so i thought i'd ask a question:

Paul has 8 cards.
There is a number on each card.

2,3,3,4,5,5,5,5

Paul takes at random 3 cards.
He adds together the 3 numbers on the card to get a total T

Work out the probability that T is an odd number.

Thanks in advance.

One of the keys is to realize that the actual numbers on the cards do not matter. The only thing that IS important is whether the cards are even or odd. There are 6 odd cards and 2 even cards. Now since we are choosing 3 cards at random and adding the cards together, the ORDER that we choose the cards ALSO DOES NOT matter. This means this is a combination problem. So how many ways can we choose 3 cards out of 8? 8C3 = 56 different ways.

In order to have an odd total there are two different 3-card hands that will accomplish this:

1.

3 odd cards. So to figure out how many ways this can occur, we calculate how many ways we can choose 3 cards from among the 6 odd cards--6C3 ways or 20 ways.

2.

2 even cards + 1 odd card. So how many ways can we choose 2 even cards out of 2 even cards--2C2 = one way and then we calculate the number of ways we can 1 odd card from the remaining 6 odd cards--6C1 = 6 ways. So, the total number of ways to get this type of 3-card hand is 1*6 = 6 ways.

3.

So the total number of ways of ending with an odd total is 20 + 6 or 26 ways.

Finally, to find the probability of choosing 3 cards at random and having an odd total, we must divide the total number of ways of having an odd total by the total number of ways of choosing 3 cards out of 8.

26 ways of total odd 13
____________________ = ___________ or about 46%
56 total ways of choosing 3 cards from 8 28