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

 It is currently 14 Oct 2019, 18:50

### 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 game of archery, a person wins if he hits the target, and .......

Author Message
TAGS:

### Hide Tags

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3074
In a game of archery, a person wins if he hits the target, and .......  [#permalink]

### Show Tags

Updated on: 30 Jan 2019, 11:39
00:00

Difficulty:

45% (medium)

Question Stats:

58% (02:56) correct 42% (02:44) wrong based on 36 sessions

### HideShow timer Statistics

In a game of archery, a person wins if he hits the target, and loses if he misses the target. The probability that A wins is $$\frac{1}{6}$$, the probability that B loses is $$\frac{2}{5}$$, and the probability that C wins is $$\frac{3}{4}$$. Find the probability that at least two of them loses the game.

A. $$\frac{3}{40}$$

B. $$\frac{7}{30}$$

C. $$\frac{7}{15}$$

D. $$\frac{19}{40}$$

E. $$\frac{1}{2}$$

_________________

Originally posted by EgmatQuantExpert on 30 Jan 2019, 01:40.
Last edited by EgmatQuantExpert on 30 Jan 2019, 11:39, edited 1 time in total.
NUS School Moderator
Joined: 18 Jul 2018
Posts: 1026
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

### Show Tags

30 Jan 2019, 07:14
The probability of A winning = 1/6, A losing = 5/6
The probability of B winning = 3/5, B losing = 2/5
The probability of C winning = 3/4, C losing = 1/4

A-winning; A'-Losing

At least 2 losing, = AB'C'+A'BC'+A'B'C+A'B'C' = 2/5*1/4*1/6+5/6*3/5*1/4+5/6*2/5*3/4+5/6*2/5*1/4 = 57/120 = 19/40

chetan2u sir, Where have I made mistake?
_________________
Press +1 Kudos If my post helps!
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4987
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

### Show Tags

30 Jan 2019, 07:17
given
A= win ; 1/6 loss 5/6
B= win; 3/5;loss 2/5
C=win; 3/4; loss 1/4

P that atleast 2 of them lose game
total 4 cases
case 1 ; A wins , B & C lose; 1/6*2/5*1/4= 2/120
case 2 A lose, B wins & C lose ; 5/6*3/5*1/4 = 15/120
case 3 A & B lose B wins ; 5/6*2/5*3/4 = 30/120
case 4 ; all of them lose ; 5/6*2/5*1/4 = 10/120

total = (2+15+30+10)/120 ; 57/120 or say 19/40

Bunuel & chetan2u ; I am not sure where am I going wrong here,

EgmatQuantExpert wrote:
In a game of archery, a person wins if he hits the target, and loses if he misses the target. The probability that A wins is $$\frac{1}{6}$$, the probability that B loses is $$\frac{2}{5}$$, and the probability that C wins is $$\frac{3}{4}$$. Find the probability that at least two of them loses the game.

A. $$\frac{3}{40}$$

B. $$\frac{7}{30}$$

C. $$\frac{7}{15}$$

D. $$\frac{1}{2}$$

E. $$\frac{3}{4}$$

_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4987
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

### Show Tags

30 Jan 2019, 07:19
Afc0892 wrote:
The probability of A winning = 1/6, A losing = 5/6
The probability of B winning = 3/5, B losing = 2/5
The probability of C winning = 3/4, C losing = 1/4

A-winning; A'-Losing

At least 2 losing, = AB'C'+A'BC'+A'B'C+A'B'C' = 2/5*1/4*1/6+5/6*3/5*1/4+5/6*2/5*3/4+5/6*2/5*1/4 = 57/120 = 19/40

chetan2u sir, Where have I made mistake?

Afc0892
Hi same issue with my solution.. seems something wrong with either question or answer choice
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
NUS School Moderator
Joined: 18 Jul 2018
Posts: 1026
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

### Show Tags

30 Jan 2019, 07:22
Archit3110 wrote:
Afc0892 wrote:
The probability of A winning = 1/6, A losing = 5/6
The probability of B winning = 3/5, B losing = 2/5
The probability of C winning = 3/4, C losing = 1/4

A-winning; A'-Losing

At least 2 losing, = AB'C'+A'BC'+A'B'C+A'B'C' = 2/5*1/4*1/6+5/6*3/5*1/4+5/6*2/5*3/4+5/6*2/5*1/4 = 57/120 = 19/40

chetan2u sir, Where have I made mistake?

Afc0892
Hi same issue with my solution.. seems something wrong with either question or answer choice

Archit3110, Hope it is. I had solved this in my office, but my answer was in none of the answer options. had saved my solution in notepad and posted it now.
_________________
Press +1 Kudos If my post helps!
Math Expert
Joined: 02 Aug 2009
Posts: 7955
Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

### Show Tags

30 Jan 2019, 09:23
Afc0892 wrote:
The probability of A winning = 1/6, A losing = 5/6
The probability of B winning = 3/5, B losing = 2/5
The probability of C winning = 3/4, C losing = 1/4

A-winning; A'-Losing

At least 2 losing, = AB'C'+A'BC'+A'B'C+A'B'C' = 2/5*1/4*1/6+5/6*3/5*1/4+5/6*2/5*3/4+5/6*2/5*1/4 = 57/120 = 19/40

chetan2u sir, Where have I made mistake?

Let us do opposite..
Atleast 2 lose means opposite of 1 loses or none..
None = all wins = (1/6)*(3/5)*(3/4)=3/40
1 loss, so two wins..= (1/6)*(2/5)*(3/4)+(1/6)*(3/5)*(1/4)+(5/6)*(3/5)*(3/4)=1/20+1/40+3/8=18/40..
So total 18/40+3/40=21/40..
Again atleast 2 lose = 1-(21/40)=19/40

_________________
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3074
Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

### Show Tags

30 Jan 2019, 11:48
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3074
In a game of archery, a person wins if he hits the target, and .......  [#permalink]

### Show Tags

01 Feb 2019, 01:55

Solution

Given:
• The probability that A wins = $$\frac{1}{6}$$
• The probability that B loses = $$\frac{2}{5}$$
• The probability that C wins = $$\frac{3}{4}$$

To find:
• The probability that at least two of them lose the game

Approach and Working:
P(atleast two lose the game) = P(all three lose) + P(exactly one wins)
• P(exactly one wins) =$$\frac{1}{6} * \frac{2}{5} * (1 - \frac{3}{4}) + (1 – \frac{1}{6}) * \frac{2}{5} * \frac{3}{4}+ (1 - \frac{1}{6}) * (1 – \frac{2}{5}) * (1 – \frac{3}{4}) = \frac{1}{60} + \frac{1}{4} + \frac{1}{8} = \frac{47}{120}$$

• P(all three lose) = $$\frac{5}{6} * \frac{2}{5} * \frac{1}{4} = \frac{1}{12}$$

Therefore, P(at least two lose the game) = $$\frac{47}{120} + \frac{1}{12} = \frac{57}{120} = \frac{19}{40}$$

Hence the correct answer is Option D.

_________________
In a game of archery, a person wins if he hits the target, and .......   [#permalink] 01 Feb 2019, 01:55
Display posts from previous: Sort by