We love probability!! (1 Viewer)

[Lazarus]

Q 7(c) of the Cranbrook 2002 trial:

Consider a pack of 50 playing cards that consists of 5 colours (yellow, green, blue, indigo and violet), each of which contain cards numbered from 1 to 10 inclusive, respectively.

A joker is added to the pack. The joker can stand for any card and when there are equal numbers of different cards it takes the value of the higher card.

Otherwise, the joker stands for the card which is occurring most often.

e.g. two 4s, two 6s and a joker = two 4s and three 6s;
or two 4s, one 6, one 8 and a joker = three 4s, one 6 and one 8.

If five cards are dealt to a player, determine the probability (leaving your answer as a fraction in simplest form) that the player has received:

(i) four 10s

(ii) any three of one number and any two of another number, e.g. three 10s and two 8s


The answer we came up with for (i) is 9/156604, which may or may not be correct. Spice girl or others want to have a go?
And the answers we came up with for (ii) were 3/4606 and 300/39151, depending on the method.

 

[BlackJack]

There are 51 cards. I believe the joker would be identical to the number it impersonates, eg. '10'
i) Two possibilities:
four real tens and a non-joker card.
5C4 * 45C1
5*45...
three real tens and a joker, and one other card,
5C3 * 1 * 45C1
10 * 45

15*45 = 675
Divide by all possible ways 51C5
= 675/2349060
=45/156604...

ii) in a moment...
edit: part ii now available in all good forums.
again, two possible compositions.
All of them authentic:
5C3 * 10 * 5C2 * 9
=9000
2 of each and a joker:
5C2 * 10 * 5C2 * 9 / 2
= 4500
P:
=13500/2349060
=225/39151

Hmm... that's some conflicting stuff...

 

[Bambul]

I started trying to get the answer. After looking back at my working and wondering if I had taken the right method I remembered: "Hey, I did my 4unit HSC last year, I don't have to do this!"

So I stopped.

 

[McLake]

Q1:

I: 4 real + 1 card
5/51 * 4/50 * 3/49 * 2/48 * 1 = 1/49980

II: 3 real + 1 joker + 1 card
5/51 * 4/50 * 3/49 * 1/48 * 1 = 1/99960

I + II = 1/13320

THIS IS'NT RIGHT, IS IT ....

Q2:

AHHHHHHHHHHH!!!!!!!

SO, what was Q8 like?

 

[BlackJack]

Well, it seems like we have another here who voluntarily came back after the hsc... this forum is that good!?

 

[martin]

Agree with blackjack about part i)
Here's my go at part ii)

If you have 2A's and 3B's (A,B arbitrary)
either

A > B
2A's 2B's one joker
hands = 5C2 * 5C2 * 1 = 100
3A's 2B's
hands = 5C3 * 5C2 = 100

OR

A < B
3A's 2B's
hands = 5C3 * 5C2 = 100
(can't have joker because A < B so joker would become B)

Now, for A > B
combinations of AB are 10,9 10,8 ..... 9,8 9,7 ......... 2,1
combinations = 8 + 7 + 6 +5 +4 +3 +2 +1 = 36

for A < B
combinations are 9,10 8,9 8,10 ..... 1,10
combinations = 1 + 2 + 3 + 4 +5 +6 +7 +8 + 9 = 45

therefore, total number of hands = 36 * (100 + 100) + 45 * 100 =11200

P (3 of one kind, 2 another) = 11200 / 51C5 = 195 / 39151

Well I have no idea if that's right or not but I think it sounds good.

Good luck tomorrow,
Martin