S1

Hi,

I need help with a few more questions on discrete random variables (again!) and hope someone can explain how to get the answers.

1) A darts player practices throwing a dart at the bull's-eye on a dart boars. Independently for each throw, her probability of hitting the bull's-eye is 0.2. Let X be the number of throws she makes, up to and including her first success.

a) Write down an equation for the probability distribution of X.

I managed to work this part out and got P(X=x) = 0.2(0.8)^(x-1), which is 0.2 times 0.8 to the power of x-1 incase it isn't clear. This is the correct answer

b) Find the probability that she will have atleast 3 failures before her first success.

I'm not sure how you do this, but the answer is 0.512

2) Six fuses, of which two are defective and four are good, are to be tested one after another in random order until both defective fuses are identified. Find the probability that the number of fuses that will be tested is:

a) Three
b) four or more

I'm not sure how to do this either. I was thinking that you can do it using probability by writing down all the combinations, but is there any other way, as this is part of discrete random variables, so maybe you have to use something to do with that instead?

The answers are a)2/15, b)7/15

Thank you very much for your help!

Reply 1

Jonny W

8

(1b)
P(at least three failures before first success)
= P(X >= 4)
= 1 - P(X <= 3)
= 1 - [P(X]
= 0.512

I think you have mistyped (2b) or its answer. Please check.

Reply 2

samd

10

2.
(a) So P(three fuses are tested before both are found) = 4/6 x 2/5 x 1/4 x 2 = 2/15
Why is it that? Consider the tree diagram attached. There are two roots my which the event can occur, and both have the same overall probabilty. I have used one of them and then multiplied by two.

treediagram.gif12.1KB

Reply 3

sweet_gurl

OP

4

Thank you both very much for your help.

For 2)b) I did P(X>=4) = 1-P(X<=3) where X is the number of fuses that are tested.

= 1 - [P(X]
= 1 - [(2/6*1/5)+(2/15)]
= 1 - [2/30+2/15]
= 1 - [1/15+2/15]
= 12/15

However the book gives the answer 7/15. Have I gone wrong somewhere?

Reply 4

Jack

2.
(a) So P(three fuses are tested before both are found) = 4/6 x 2/5 x 1/4 x 2 = 2/15
Why is it that? Consider the tree diagram attached. There are two roots my which the event can occur, and both have the same overall probabilty. I have used one of them and then multiplied by two.

Surely you can go down, GFG, GGF or FGG

Reply 5

samd

10

If this is the exercise that I am thinking of (i.e. Ex BA in Heinmann's S1 Book Q15) then I think that you'll find the question asks for P(4 or LESS fuses are tested). You have found P(4 or MORE).

I don't have a copy of the book any more. I am inferring this from my working. It may be that I misunderstood the question myself at the time.

Check the question.

Reply 6

sweet_gurl

OP

4

samd

If this is the exercise that I am thinking of (i.e. Ex BA in Heinmann's S1 Book Q15) then I think that you'll find the question asks for P(4 or LESS fuses are tested). You have found P(4 or MORE).

I don't have a copy of the book any more. I am inferring this from my working. It may be that I misunderstood the question myself at the time.

Check the question.

Yes, you're right, I had copied down the question wrong and then I was working from the wrong answer rather than the book! :redface:

I just did it as 4 or fewer as the question asked, and got the right answer. Thanks! :smile:

S1

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