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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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New post Updated on: 30 Jan 2019, 11:39
00:00
A
B
C
D
E

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  45% (medium)

Question Stats:

58% (02:56) correct 42% (02:44) wrong based on 36 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, 01:40.
Last edited by EgmatQuantExpert on 30 Jan 2019, 11:39, 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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New post 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?
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Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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New post 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 please check question and answer option



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

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

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New post 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?



I don't find anything wrong, and your answer is correct..

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

So you are perfectly fine with your answer.
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Re: In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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New post 30 Jan 2019, 11:48
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In a game of archery, a person wins if he hits the target, and .......  [#permalink]

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New post 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.

Answer: D

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