Probability

jamo2oo9

Hi all,

I've been having trouble with Bernoulli trial questions that would be on the Leaving Cert paper. Could anyone advise me what am I supposed to do in the following question please

'A certain basketball player scores 60% of the free-throw shots she attempts. During a particular game, she gets 6 free throws.'

Question (a) asked for points needed to make a Bernoulli trial work. I got that no problem.

Question (b) asks us to find, correct to 3 decimal places, the probability that she scores on exactly four of the six shots

My answer for that was 0.329 using the formula I use 'R' instead of K in the formula as we were taught with that.

The second part of the question asked me what is the probability that she scores for the second time on the fifth shot. I don't understand this bit. Could someone help with this part please?

Cheers,
Jamie

MathsManiac

If she scores for the second time on the fifth shot, that means that she scores exactly once among the first four shots, and then scores on the fifth shot.

So, its P(one success in four trials) X P(success in one trial).

By the way, check your calculations for part (b)(i), I think you're slightly out.

Also, this is from last year's exam paper, so if you want to check your answers, you can check the marking scheme. It's on P44 of this document: http://www.examinations.ie/archive/markingschemes/2012/LC003ALP030EV.pdf

jamo2oo9

Ah cheers for that mate! Used the wrong values for the (b) (i) but got it fixed now.

Leaving Cert Student

I am struggling with this probability problem:

Mary is playing cards with Liam and Sam. She has nine clubs in her hand. The other fourn clubs, A, 2, 3, 4 are randomly dealt to Liam and Sam. Liam holds at least 2 of the 4 cards and may have all 4.

if Liam shows Mary that he has the 2 of clubs, what is the probability that Sam has the ace of clubs?

I keep getting 1/3 but answer is 3/7 somehow?

MathsManiac

To avoid getting into a debate like the Monty Hall one, let's assume that instead of "If Liam shows Mary that he has the 2 of clubs,..." we have: "Mary asks Liam whether he has the 2 of clubs, and he says yes." (If you don't understand why a lot of people think this matters, then ignore it.)

Anyway, with that out of the way, there are seven possibilities in total, which we are assuming are equally likely:
Liam has A,2 and Sam has 3,4
Liam has 2,3 and Sam has A,4
Liam has 2,4 and Sam has A,3
Liam has A,2,3 and Sam has 4
Liam has A,2,4 and Sam has 3
Liam has 2,3,4 and Sam has A
Liam has A,2,3,4 and Sam has none.

In three of these, Sam has the ace.

Leaving Cert Student

MathsManiac wrote: »

To avoid getting into a debate like the Monty Hall one, let's assume that instead of "If Liam shows Mary that he has the 2 of clubs,..." we have: "Mary asks Liam whether he has the 2 of clubs, and he says yes." (If you don't understand why a lot of people think this matters, then ignore it.)

Anyway, with that out of the way, there are seven possibilities in total, which we are assuming are equally likely:
Liam has A,2 and Sam has 3,4
Liam has 2,3 and Sam has A,4
Liam has 2,4 and Sam has A,3
Liam has A,2,3 and Sam has 4
Liam has A,2,4 and Sam has 3
Liam has 2,3,4 and Sam has A
Liam has A,2,3,4 and Sam has none.

In three of these, Sam has the ace.

cheers, i get what you mean about the monTy python problem..
I had been looking at each arrangement (ie. liam has 2 cards or 3 or 4) as equally likely not each specific card assortment. thanks