GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 14 Oct 2019, 18:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# In a shooting competition, probability of A hitting target is 2/5, by

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 58320
In a shooting competition, probability of A hitting target is 2/5, by  [#permalink]

### Show Tags

13 Nov 2018, 10:48
00:00

Difficulty:

25% (medium)

Question Stats:

75% (01:59) correct 25% (01:53) wrong based on 82 sessions

### HideShow timer Statistics

In a shooting competition, probability of A hitting target is 2/5, by B is 2/3 and C is 3/5. If all of them fire independently, what is the probability that only one will hit target?

(A) 3/25
(B) 4/25
(C) 1/3
(D) 2/3
(E) 3/4

_________________
Manager
Joined: 09 Jun 2018
Posts: 188
Location: United States
GPA: 3.95
WE: Manufacturing and Production (Energy and Utilities)
Re: In a shooting competition, probability of A hitting target is 2/5, by  [#permalink]

### Show Tags

13 Nov 2018, 11:07
1
2
P(x) = probability of hitting target, P(x") = probability of not hitting target = 1 - P(x), using this:

P(A) = 2/5, so, P(A") = 3/5
P(B) = 2/3, so, P(B") = 1/3
P(C) = 3/5, so, P(C") = 2/5

OR = plus, and = *

We need: P(A hits & B doesnt & C doesnt) OR P(A doesnt & B hits & C doesnt) OR P(A doesnt & B doesnt & C hits)
which is: P(A)(B")(C") + P(A")P(B)P(C") + P(A")P(B")P(C)

= $$(2/5)*(1/3)*(2/5) + (3/5)*(2/3)*(2/5) + (3/5)*(1/3)*(3/5)$$

= $$\frac{(4+12+9)}{75}$$

= 25/75 = 1/3

Option C
_________________
If you found this relevant and useful, please Smash that Kudos button!
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4987
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In a shooting competition, probability of A hitting target is 2/5, by  [#permalink]

### Show Tags

14 Nov 2018, 02:10
Bunuel wrote:
In a shooting competition, probability of A hitting target is 2/5, by B is 2/3 and C is 3/5. If all of them fire independently, what is the probability that only one will hit target?

(A) 3/25
(B) 4/25
(C) 1/3
(D) 2/3
(E) 3/4

P (A)= 2/5 , Pn(A)= 3/5
P(B)= 2/3, Pn(B)=1/3
P(C)= 3/5, Pn(C)=2/5

only one will hit Probability=
{2/5*1/3*2/5}+{3/5*2/3*2/5}+{3/5*1/3*3/5}

= 25/75
=1/3 option C
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8043
Location: United States (CA)
Re: In a shooting competition, probability of A hitting target is 2/5, by  [#permalink]

### Show Tags

27 Jan 2019, 21:04
Bunuel wrote:
In a shooting competition, probability of A hitting target is 2/5, by B is 2/3 and C is 3/5. If all of them fire independently, what is the probability that only one will hit target?

(A) 3/25
(B) 4/25
(C) 1/3
(D) 2/3
(E) 3/4

P(A hits and B,C miss) = 2/5 x 1/3 x 2/5 = 4/75

P(B hits and A, C miss) = 3/5 x 2/3 x 2/5 = 12/75

P(C hits and A, B miss) = 3/5 x 1/3 x 3/5 = 9/75

So the total probability is 4/75 + 12/75 + 9/75 = 25/75 = 1/3.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: In a shooting competition, probability of A hitting target is 2/5, by   [#permalink] 27 Jan 2019, 21:04
Display posts from previous: Sort by