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

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In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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Updated on: 30 Jan 2019, 10:39
00:00

Difficulty:

65% (hard)

Question Stats:

38% (03:02) correct 62% (02:26) wrong based on 25 sessions

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

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Originally posted by EgmatQuantExpert on 30 Jan 2019, 00:40.
Last edited by EgmatQuantExpert on 30 Jan 2019, 10:39, edited 1 time in total.
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Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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30 Jan 2019, 06:14
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?
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Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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30 Jan 2019, 06:17
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. $$\frac{3}{4}$$

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Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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30 Jan 2019, 06:19
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
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Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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30 Jan 2019, 06:22
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.
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Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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30 Jan 2019, 08:23
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

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Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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30 Jan 2019, 10:48
Hello,

There was a small mistake in the approach. We have changed the options now.

Regards,
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In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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01 Feb 2019, 00:55

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 C.

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In a game of archery, a person wins if he hits the target, and .......   [#permalink] 01 Feb 2019, 00:55
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