Cambridge Combinations Question (1 Viewer)

[Kyrix]

Not sure how to obtain the answer to this probability question in the cambridge text book.

A poker hand of five cards is dealt from a standard pack of 52. Find the probability of obtaining
(a) one pair (answer: 352/833)
(b) two pairs (answer: 198/4165)
(c) three of a kind
(d) four of a kind

Its on page 437 question 22 in the year 12 3u cambridge if anyone is interested

 

[Kyrix]

Ok.. since no one is answering =/

Do you think questions in the hsc will get this hard?

Because imo the hsc past paper combinations are relatively easy compared to this

 

[OldMathsGuy]

I think what this question highlights is the uncertainty (pardon the pun) that students have in knowing how to approach these sorts of questions. Next year when they go to multiple choice style questions for the first part, I think this type of question could seriously confuse over half the candidature.

The following should help you think about these the correct way:

You need to choose a card value to pair up: 13C1 multiply by
Of which you choose 2 out of the possible 4 suits available for that value: 4C2 multiply by
Of which you then need to choose 3 values from the other 12 values: 12C3 multiply by
Of which you then need to choose 1 out of the 4 suits available for each of these cards: 4C1 x 4C1 x 4C1

Which is 1 098 240 different ways of picking 5 cards but only a single pair among them.
In terms of probability there are 52C5 ways of choosing 5 cards: that is 2 598 960 ways.

Divide the first by the latter to get 352/833.

It is a similar process for the other parts (give it a go and see how you fair).

Once you get the hang of these questions and how to frame them, they become routine but until you get to that point, they confuse a lot of students.
I doubt you would see this style of question on the Ext 1 exam.

Best Regards
OldMathsGuy

 

[muzeikchun852]

(b) 4C2 x 4C2 x 13C2 x 44C1 / 52C5 = ANSWER.
(c) 4C3 x 13 x 48C2 / 52C5 = 94/4165
(d) 4C4 x 13 x 48C1 / 52C5 = 1/4165

 

[Kyrix]

I think what this question highlights is the uncertainty (pardon the pun) that students have in knowing how to approach these sorts of questions. Next year when they go to multiple choice style questions for the first part, I think this type of question could seriously confuse over half the candidature.

The following should help you think about these the correct way:

You need to choose a card value to pair up: 13C1 multiply by
Of which you choose 2 out of the possible 4 suits available for that value: 4C2 multiply by
Of which you then need to choose 3 values from the other 12 values: 12C3 multiply by
Of which you then need to choose 1 out of the 4 suits available for each of these cards: 4C1 x 4C1 x 4C1

Which is 1 098 240 different ways of picking 5 cards but only a single pair among them.
In terms of probability there are 52C5 ways of choosing 5 cards: that is 2 598 960 ways.

Divide the first by the latter to get 352/833.

It is a similar process for the other parts (give it a go and see how you fair).

Once you get the hang of these questions and how to frame them, they become routine but until you get to that point, they confuse a lot of students.
I doubt you would see this style of question on the Ext 1 exam.

Best Regards
OldMathsGuy

Wow thanks.

I sort of get it now.

Combinations always makes my head spin =/