Answer? :s (1 Viewer)

[rr.dun.dun]

:mad1:. .

How do you do this? I always can't figure how the simplest question.

(PROBABILITY)
From a normal deck of cards, 4 cards are randomly selected. Calculate the probability that the hand contains:

a) 4 queens
b) an ace, king, queen and jack
c) the ace, king, queen and jack of clubs
d) one 5
e) three 10s and an ace
f) 2 red cards and 2 black cards

Please answer it, it is pretty simple but i got stuck on one so i stopped trying.

Thank you!

 

[PC]

First of all ... is this a permutation or a combination question? Is the ORDER in which the cards are dealt important? No. So it's a combination question.

Next, work out the total number of possible hands that could be dealt from a deck of cards.
Number of hands = (52 x 51 x 50 x 49) ÷ (4 x 3 x 2 x 1)
= 270725

This number will be the denominator of all the probability questions.

Now try each question.

(a) For 4 queens, you'll need all 4 queens.
To get the queens, number of ways = (4 x 3 x 2 x 1) ÷ (4 x 3 x 2 x 1) = 1
So P(4 queens) = 1/270725

(b) For A, K, Q, J, we'll need any Ace AND any King AND any Queen AND any Jack.
For the Ace, number of ways = 4 ÷ 1 = 4
For the King, number of ways = 4 ÷ 1 = 4
For the Queen, number of ways = 4 ÷ 1 = 4
For the Jack, number of ways = 4 ÷ 1 = 4
So, number of ways = 4 x 4 x 4 x 4 = 256
So, P(AKQJ) = 256/270725

(c) We need the Ace of Clubs AND the King of Clubs AND the Queen of Clubs AND the Jack of Clubs.
For the Ace of Clubs, number of ways = 1 ÷ 1 = 1
For the King of Clubs, number of ways = 1 ÷ 1 = 1
For the Queen of Clubs, number of ways = 1 ÷ 1 = 1
For the Jack of Clubs, number of ways = 1 ÷ 1 = 1
So, number of ways = 1 x 1 x 1 x 1 = 1
So, P(AKQJclubs) = 1/270725

(d) We need one 5 of any suit, then any three other cards.
For the 5, number of ways = 4 ÷ 1 = 4
For the other cards, number of ways = (51 x 50 x 49) ÷ (3 x 2 x 1) = 20825
So, number of ways = 4 x 20825 = 83300
So, P(one 5) = 83300/270725 = 4/13
(which kind of makes sense anyway)

OR if this question means ONLY one 5, then we need one 5 of any suit, then any three other cards, provided they're not 5.
For the 5, number of ways = 4 ÷ 1 = 4
For the other cards, number of ways = (48 x 47 x 46) ÷ (3 x 2 x 1) = 17296
So, number of ways = 4 x 17296 = 69184
So, P(one 5) = 69184/270725

(e) We need any three 10s AND any ace.
For the three 10s, number of ways = (4 x 3 x 2) ÷ (3 x 2 x 1) = 4
For the Ace, number of ways = 4 ÷ 1 = 4
So, number of ways = 4 x 4 = 16
So, P(3 10s,Ace) = 16/270725

(f) We need any two red cards AND any two black cards.
For the red cards, number of ways = (26 x 25) ÷ (2 x 1) = 325
For the black cards, number of ways = (26 x 25) ÷ (2 x 1) = 325
So, number of ways = 325 x 325 = 105625
So, P(2 red, 2 black) = 105625/270725 = 325/833

Hope this helps!

 

[rr.dun.dun]

Thank you so much. I understand it now.

 

[watatank]

is probability in general maths similar to the ones you see in two unit maths? it's not an easy question...

 

[PC]

Actually this type of probability is not seen in 2 unit Maths. It's actually in the topic of permutations and combinations which is only studied in 3 unit Maths.

 

[watatank]

Actually this type of probability is not seen in 2 unit Maths. It's actually in the topic of permutations and combinations which is only studied in 3 unit Maths.

i could do the first one using two unit methods...so its like a probability without replacement. sort of. i kind of imagined i picked out one card at a time out of a deck of cards and calculated the probability of the respective cards being queens.

P(QQQQ) = (4/52) * (3/51) * (2/50) * (1/49) = 1/270725

haven't tried the others, they get quite complicated, obviuosly more complicated than that..but i dont think its beyond two unit students.

 

[PC]

There are some other types of probability questions in the General Course which 2 unit students can do in a couple of lines, but the General students need to draw a tree diagram to do it. Obviously the best General students will figure out what's going on.

 

[Shock]

Like my teacher says, "Do a tree diagram" i find it very useful