• Support PF! Buy your school textbooks, materials and every day products Here!

Probability problem

  • Thread starter twoflower
  • Start date
368
0
1. Homework Statement

Bowman shoots into a dartboard, with possible gain ranging from 0 to 10 points.

Probability that he shoots 30 points in 3 shots is 0.008.

Probability that he shoots < 8 in one shot is 0.4.

Probability that he shoots exactly 8 in one shot is 0.15.

What is the probability that he gains at least 28 points in 3 shots?

3. The Attempt at a Solution

My solution:

[tex]
P(X \ge 28) = P(X = 28) + P(X = 29) + P(X = 30)
[/tex]

We know P(X = 30) so it's sufficient to count P(X = 28) and P(X = 29).

X = 28

This situation can occur either if:
(a) He shoots 10, 10 and 8 (in any order)
(b) He shoots 10, 9, 9 (in any order)

So I guess:

[tex]
(*)\ \ \ \ P(X = 28) = (P(X = 10).P(X =10).P(X = 8))+(P(X=10).P(X=9).P(X=9))
[/tex]

What I'm interested in is whether this is ok. I don't know if I should take into an account that (let's take for example the case (a)) the bowman can shoot the points in any order, ie. 10, 10, 8 or 10, 8, 10 and so on.

Don't I have to multiply (*) with 3! so that I cover all the orders in which the shooter can gain those points?

Thank you very much.
 

Answers and Replies

D H
Staff Emeritus
Science Advisor
Insights Author
15,329
681
So I guess:

[tex]
(*)\ \ \ \ P(X = 28) = (P(X = 10).P(X =10).P(X = 8))+(P(X=10).P(X=9).P(X=9))
[/tex]

What I'm interested in is whether this is ok. I don't know if I should take into an account that (let's take for example the case (a)) the bowman can shoot the points in any order, ie. 10, 10, 8 or 10, 8, 10 and so on.
Yes, you have to account for the fact that Bowman can shoot a 28 by any combination of 10,10,8 or 10,9,9.

Don't I have to multiply (*) with 3! so that I cover all the orders in which the shooter can gain those points?
Why 3!=6 ? How many distinct permutations of 10,10,8 exist? (Hint: Not 6).

You also need to determine the probabilities that Bowman will shoot exactly 9 and exactly 10 in a single shot to answer the question.
 
368
0
Yes, you have to account for the fact that Bowman can shoot a 28 by any combination of 10,10,8 or 10,9,9.

Why 3!=6 ? How many distinct permutations of 10,10,8 exist? (Hint: Not 6).
I'm little confused, why shouldn't I distinguish between those two 10-points shots?

Ok, if I won't distinguish between them, then I will have 3 ways of ordering the shots.

And, is there any simple answer to WHY do I have to take the order of the shots into account? I just can't see it...


You also need to determine the probabilities that Bowman will shoot exactly 9 and exactly 10 in a single shot to answer the question.
I did it this way:

[tex]
P(X = 10) = \sqrt[3]{P(X=30)}
[/tex]

[tex]
P(X \le 9) = 1 - P(X = 10)
[/tex]

[tex]
P(X = 9) = P(X \le 9) - (P(X < 8) + P(X = 8))
[/tex]
 
D H
Staff Emeritus
Science Advisor
Insights Author
15,329
681
Twoflowers, I am answering your reply in reverse order.

You also need to determine the probabilities that Bowman will shoot exactly 9 and exactly 10 in a single shot to answer the question.
I did it this way:
[tex]
P(X = 10) = \sqrt[3]{P(X=30)}
[/tex]

[tex]
P(X \le 9) = 1 - P(X = 10)
[/tex]

[tex]
P(X = 9) = P(X \le 9) - (P(X < 8) + P(X = 8))
[/tex]
Correct.

Why 3!=6 ? How many distinct permutations of 10,10,8 exist? (Hint: Not 6).
I'm little confused, why shouldn't I distinguish between those two 10-points shots?

Ok, if I won't distinguish between them, then I will have 3 ways of ordering the shots.

And, is there any simple answer to WHY do I have to take the order of the shots into account? I just can't see it...

Note that you were able to calculate [itex]P(X = 10) = \sqrt[3]{P(X=30)}
[/itex] because there is only one way to get 30: 10 on each shot. You did not distinguish between the 10s because they are indistinguishable. Similarly, there are three ways (not six) to shoot a 28 by shooting two tens and an eight: 10,10,8; 10,8,10; 8,10,10. These three combinations are distinguishable. On the other hand, 10,10,8 is indistinguishable from 10,10,8.

Switch gears to a different problem: Rolling a pair of dice. There are 36 total combinations. There is only one way to get boxcars: probability=1/36. There are two ways to get a total of eleven: 5+6 or 6+5. Probability=2/36. You have to account for the order because the two events are distinct events that happen to have same outcome as far as the total is concerned.
 
368
0
Note that you were able to calculate [itex]P(X = 10) = \sqrt[3]{P(X=30)}[/itex] because there is only one way to get 30: 10 on each shot. You did not distinguish between the 10s because they are indistinguishable. Similarly, there are three ways (not six) to shoot a 28 by shooting two tens and an eight: 10,10,8; 10,8,10; 8,10,10. These three combinations are distinguishable. On the other hand, 10,10,8 is indistinguishable from 10,10,8.

Switch gears to a different problem: Rolling a pair of dice. There are 36 total combinations. There is only one way to get boxcars: probability=1/36. There are two ways to get a total of eleven: 5+6 or 6+5. Probability=2/36. You have to account for the order because the two events are distinct events that happen to have same outcome as far as the total is concerned.
Thank you DH, now it's much more clear to me :smile:
 

Related Threads for: Probability problem

  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
8
Views
4K
  • Last Post
Replies
13
Views
2K
Top