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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: kriptik on September 12, 2010, 10:01:10 pm

Title: another probability question
Post by: kriptik on September 12, 2010, 10:01:10 pm

Im stuck on this question , anyone that can help :D

Title: Re: another probability question
Post by: cypriottiger on September 12, 2010, 10:13:21 pm

im not the best at methods but i had a try :)
i used the equation (n *over* x)p^x X q^n-x
so your chance of success or Pr(S) is 0.25 or p, and your chance of failure or Pr(F) is 0.75 or q
you need Pr (x < or equal to 2)
so number of opportunities (n) in the calculator with the nCr( function is (5,2) X 0.25^2 X 0.75^3 = 135/512. but thats the Pr(X=2) you do the same for X=1 and X=0, and i got the answer 459/512 or 0.8965 :)

Title: Re: another probability question
Post by: sb3700 on September 12, 2010, 10:14:01 pm

We want to either get 1 heart or 0 hearts.

Let X be the number of hearts.

Pr(X=0) = 39*38*37*36*35 / (52*51*50*49*49)

To find Pr(X=1), need to consider the ways of drawing exactly 1 heart:
HNNNN, NHNNN, ..., NNNNH, ncr(5, 1) = 5 ways in all

and the probability of each way is the same:
Pr(HNNNN) = 13 * 39 * 38 * 37 * 36 / (52*51*50*49*49)
(make sure you can see why).

So Pr(X = 1) = 5 * 13 * 39 * 38 * 37 * 36 / (52*51*50*49*49)

and Pr(X < 2) = Pr(X=1) + Pr(X=0) = ...

(sorry, no calculator on hand)


EDIT: @cypriottiger
The probability changes each time as cards aren't replaced.

EDIT: and I forgot Pr(X=2) as it's actually Pr(X<=2)

The number of ways to get 2 hearts is nCr(5, 2) = 10
Pr(HHNNN) = 13 * 12 * 39 * 38 * 37 / (52*51*50*49*49)

so Pr(X = 2) = 10 * 13 * 12 * 39 * 38 * 37 / (52*51*50*49*49)

Title: Re: another probability question
Post by: cypriottiger on September 12, 2010, 10:16:19 pm

Quote from: sb3700 on September 12, 2010, 10:14:01 pm

We want to either get 1 heart or 0 hearts.

Let X be the number of hearts.

Pr(X=0) = 39*38*37*36*35 / (52*51*50*49*49)

To find Pr(X=1), need to consider the ways of drawing exactly 1 heart:
HNNNN, NHNNN, ..., NNNNH, ncr(5, 1) = 5 ways in all

and the probability of each way is the same:
Pr(HNNNN) = 13 * 39 * 38 * 37 * 36 / (52*51*50*49*49)
(make sure you can see why).

EDIT: @cypriottiger
The probability changes each time as cards aren't replaced.

So Pr(X = 1) = 5 * 13 * 39 * 38 * 37 * 36 / (52*51*50*49*49)

and Pr(X < 2) = Pr(X=1) + Pr(X=0) = ...

(sorry, no calculator on hand)

ah my mistake, thanks for correcting

Title: Re: another probability question
Post by: 8039 on September 13, 2010, 10:29:42 pm

You could do a tree diagram for this if you wanted to... would take at least 15 minutes tho imo