Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 19 May 2013, 04:08

# Probability-Bowman

Author Message
TAGS:
Manager
Joined: 01 Jan 2005
Posts: 169
Location: NJ
Followers: 1

Kudos [?]: 0 [0], given: 0

Probability-Bowman [#permalink]  10 Mar 2005, 06:53
Here is a simple question

A bowman hits his target in 1/5 of his shots. What is the probability of him missing the target atleast once in 4 shots.

I know how to solve this...

I saw this Binomial Distribution explained by Hong in the following thread
http://www.gmatclub.com/phpbb/viewtopic.php?t=14706

Can anybody explain me how can I solve the above question using the concept explained in the thread...

Hong, Plz help..

Thanks,
Vijo.
 Kaplan Promo Code Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes
SVP
Joined: 30 Sep 2004
Posts: 1548
Location: Germany
Followers: 4

Kudos [?]: 15 [0], given: 0

Probability that one target is missed + probability that two targets are missed + ... + Probability that four targets are missed

4c1*4/5*(1/5)^3+4c2*(4/5)^2*(1/5)^2+4c3*(4/5)^3*1/5+4c4*(4/5)^4

or

1- Probability that no targets are missed (recommended)

1 - 4c4*(1/5)^4
Intern
Joined: 19 Jul 2004
Posts: 43
Followers: 1

Kudos [?]: 3 [0], given: 0

P(Hitting) = 1/5
P(Missing) = 4/5

P(Atleast once in four shots) = 1 - P(None in four shots)

With four shots there are 16 ways out of which there is only one way in which No shots are hit. This is same as finding C(4,0) = 1

P(None in four shots) = (Number of ways this happens)(probability this happens)

= C(4,0) * (1/5)^0 * (4/5)^4
= (4/5)^4

Hence

P(Atleast once in four shots) = 1 - (4/5)^4

Ketan
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9

Kudos [?]: 157 [0], given: 0

Great job both of you!
One correction, the question says at least one miss (not at least one hit), so it is equal to 1-no miss, or 1-all hit.
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9

Kudos [?]: 157 [0], given: 0

Is 4/5 the chance of miss or the chance of hit, rupstar?
Senior Manager
Joined: 15 Feb 2005
Posts: 258
Location: Rockville
Followers: 1

Kudos [?]: 5 [0], given: 0

of what i understood 1/5 is making the target and 4/5 is missing the target

hence
him missing the target atleast once is[ 1- him never missing the target]
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9

Kudos [?]: 157 [0], given: 0

That's right. :b:

(I saw you corrected ketanm's C(4,0), thought you didn't correct the other part, which you did.)
Manager
Joined: 01 Jan 2005
Posts: 169
Location: NJ
Followers: 1

Kudos [?]: 0 [0], given: 0

Rupstar wrote:
yea 1-4c4(1/5^4)

Hi Hong

I am just trying to understand the application of the concept in that thread.

Rupstar's expression is equivalent to
1- ( C(4,4) * (1/5)^4 * (4/5)^0) )

Am I right Hong???
SVP
Joined: 03 Jan 2005
Posts: 2322
Followers: 9

Kudos [?]: 157 [0], given: 0

Yes, exactly.
Display posts from previous: Sort by