1. ## probability distribution

A crossword puzzle is published in a newspaper daily . Farah could complete , at average , 3 out of 5 crossword puzzles . Find the probability that in 4 certain weeks , Farah could complete at least 2 crossword puzzles in 3 out of the 4 weeks .

someone pls help me ..

2. Ok
Im no genius but here it goes

So bassicly he can complete 3 out of every 5 puzzles he does.
There are 4 weeks we need to work out if he can do 2 puzzles in 3 weeks so the 4 weeks isn't relevant?
now im gonna find the lowest common denominator
so this means theres a
9/15 chance he can do a puzzle
and a
10/15 chance needed to do 2/3 puzzles

that's all i can work out sorry hope i've helped

A crossword puzzle is published in a newspaper daily . Farah could complete , at average , 3 out of 5 crossword puzzles . Find the probability that in 4 certain weeks , Farah could complete at least 2 crossword puzzles in 3 out of the 4 weeks .

someone pls help me ..
This is my take on it:

The probability of completing a single crossword puzzle in any given week is 3/5 = 0.6.

Let X be the random variable number of crossword puzzles completed in a given week.

X ~ Binomial(n = 5, p = 0.6)

Calculate $P = \Pr(X \geq 2) = 1 - \Pr(X \leq 1)$.

Let Y be the random variable number of weeks in which at least two crossword puzzles are completed.

Y ~ Binomial(n = 4, p = P) (P has been calculated above).

Calculate $\Pr(Y = 3)$.

4. Originally Posted by lmcph24
Ok
Im no genius but here it goes

So bassicly he can complete 3 out of every 5 puzzles he does.
There are 4 weeks we need to work out if he can do 2 puzzles in 3 weeks so the 4 weeks isn't relevant?
now im gonna find the lowest common denominator
so this means theres a
9/15 chance he can do a puzzle
and a
10/15 chance needed to do 2/3 puzzles

that's all i can work out sorry hope i've helped
Sorry but this reply is wrong for several reasons, including

1. The fact that it's "at least 2 crossword puzzles", not 2.
2. The 4 weeks is relevant.
3. The calculated probabilities are irrelavent to the question.

5. Originally Posted by mr fantastic
Sorry but this reply is wrong for several reasons, including

1. The fact that it's "at least 2 crossword puzzles", not 2.
2. The 4 weeks is relevant.
3. The calculated probabilities are irrelavent to the question.
Im sorry for trying to help as i said
"Im no genius but here it goes"

And as you told me please don't double post
Also
if hes finding out the probability then what does 3 Pr(Y=3) mean in terms of the probability of something happen. I'm most probably wrong but i'm here to learn so please explain

6. Originally Posted by lmcph24
Im sorry for trying to help as i said
"Im no genius but here it goes"

And as you told me please don't double post
Also
if hes finding out the probability then what does 3 Pr(Y=3) mean in terms of the probability of something happen. I'm most probably wrong but i'm here to learn so please explain
If you're not sure how to do a question then don't answer it - leave it for someone who is sure. It's better for question to go unanswered for a while than for the OP to get a quick reply that is wrong.

I don't know where the following has come from or what you mean by it:
[snip]if hes finding out the probability then what does 3 Pr(Y=3) mean in terms of the probability of something happen[snip]

7. this might be really stupid but im willing to make a fool of myself.

the probability he gets 5 right is 0.07776 and he can do that in 1 way
4= 0.05184 can be done in 5 ways =0.2592
3= 0.03456 can be done in 10 ways =0.3456
2= 0.02304 can be done in 10 ways =0.2304
1= 0.01536 can be done in 5 ways =0.0768
0= 0.01024 can be done in 1 way

opps, didnt mean to send that

so Probability of two or more =0.91296

he can get 3 of 4 weeks with 2 or more (0.91296X0.91296X0.91296X0.08704) in 4 ways
= 0.2649318204

or maybe you could use binomial probability some how