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At a certain company, the probabilities of success of its three distin

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At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 22 Aug 2019, 07:09
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Competition Mode Question



At a certain company, the probabilities of success of its three distinct new product launches are independently \(\frac{1}{4}\), \(\frac{1}{2}\), and \(\frac{5}{8}\), respectively. What is the probability that exactly two of the launches succeed?

A. \(\frac{7}{8}\)

B. \(\frac{1}{2}\)

C. \(\frac{23}{64}\)

D. \(\frac{5}{64}\)

E. \(\frac{3}{64}\)

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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 22 Aug 2019, 07:17
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Probability of exactly two launches succeeded can be calculated as follows:

Launch 1: W W L
Launch 2: W L W
Launch 3: L W W

The calculation is done in the picture image below.

Answer is option C

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New post 22 Aug 2019, 07:23
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At a certain company, the probabilities of success of its three distinct new product launches are independently 1414, 1212, and 5858, respectively. What is the probability that exactly two of the launches succeed?

A. 7/8

B. 1/2

C. 23/64

D. 5/64

E. 3/64

success of a,b,c ; 1/4, 1/2 , 5/8
failure a,b,c ; 3/4, 1/2, 3/8
so P exactly two will succeed ; 1/4*1/2*3/8 + 1/2*5/8*3/4 + 1/4*5/8*1/2 ; 23/64
IMO C
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post Updated on: 22 Aug 2019, 08:39
P(A, ok)=1/4
P(A, fail)=3/4
P(B, ok)=1/2
P(B, fail)=1/2
P(C, ok)=5/8
P(C, fail)=3/8

probability that exactly two of the launches succeed =

P(A, ok) * P(B, ok) * P(C, fail) +
P(A, ok) * P(B, fail) * P(C, ok) +
P(A, fail) * P(B, ok) * P(C, ok)

=1/4 * 1/2 * 3/8 + 1/4 * 1/2 * 5/8 + 3/4 * 1/2 * 5/8

=23/64

Answer is (C)

Originally posted by chondro48 on 22 Aug 2019, 07:24.
Last edited by chondro48 on 22 Aug 2019, 08:39, edited 1 time in total.
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 22 Aug 2019, 07:32
IMO -C

Let A= 1/4
B= 1/2
C= 5/8

So exactly 2
AB not C + BC not A + AC not B
= (1/4 1/2 3/8) + (3/4 1/2 5/8) + (1/4 1/2 5/8)
=23/64

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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 22 Aug 2019, 08:19
Quote:
At a certain company, the probabilities of success of its three distinct new product launches are independently 1/4, 1/2 and 5/8 respectively. What is the probability that exactly two of the launches succeed?

A. 7/8

B. 1/2

C. 23/64

D. 5/64

E. 3/64



Given: At a certain company, the probabilities of success of its three distinct new product launches are independently 1/4, 1/2 and 5/8 respectively.

Asked: What is the probability that exactly two of the launches succeed?

The probability that exactly two of the launches succeed = \(\frac{1}{4}*\frac{1}{2}*\frac{3}{8} + \frac{1}{4}*\frac{1}{2}*\frac{5}{8} + \frac{3}{4}*\frac{1}{2}*\frac{5}{8}= \frac{(3 + 5 + 15)}{64} = \frac{23}{64}\)

IMO C
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 22 Aug 2019, 11:57
Let P(A)=1/4, then P'(A)=3/4
Let P(B)=1/2, then P'(B)=1/2
Let P(C)=5/8, then P'(C)=3/8

Probability that exactly two launches succeed, P(E)=1/4*1/2*3/8 + 1/4*5/8*1/2 +1/2*5/8*3/4
=3/64 + 5/64 + 15/64 = 23/64
Hence Answer is C.

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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 22 Aug 2019, 12:40
prob of launches 1/2, 1/4, 5/8 :
Prob launches are unsuccessful : 1/2, 3/4, 3/8 :
For exactly 2 we need to consider 3 cases, for each case 1 even shall have not successful rate: so final eq will be

1/2*1/4*3/8+ 1/2*5/8*3/4 + 1/4*5/8*1/2 = 23/64

C Answer
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 22 Aug 2019, 17:36
For this to be possible, two products have to succeed and one has to fail. Probability of failling is the rest to complete 100%. Example,if something has 30% chance of working, it will have 70% chance of not working

1 and 2 succeed, 3 fails
1/4 * 1/2 * 3/8 = 3/64

1 and 3 succeed, 2 fails
1/4 * 1/2 * 5/8 = 5/64

2 and 3 succeed, 1 fails
3/4 * 1/2 *5/8 = 15/64

The answer will then be the sum all = 23/64

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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 22 Aug 2019, 21:06
There are three ways to get exactlly two seccussful new product launches.
1) 1/4 ×1/2 × 3/8 =3/64
2) 1/4 ×5/8 × 1/2 =5/64
3) 3/4 ×5/8 × 1/2 =15/64
(3+5+15)/64 =23/64
Option C

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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 23 Aug 2019, 00:45
\(\frac{1}{4}x\frac{1}{2}x\frac{3}{8}\)
+
\(\frac{1}{4}x\frac{1}{2}x\frac{5}{8}\)
+
\(\frac{3}{4}x\frac{1}{2}x\frac{5}{8}\)
=\(\frac{3}{64}+\frac{5}{64}+\frac{15}{64}\)
=\(\frac{23}{64}\)
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 23 Aug 2019, 05:56
Three product launches (PL) = A, B, C
Probabilities of success of products: P(A) = 1/4, P(B) = 1/2, P(C)=5/8
Probability of success of exactly two product launches = P(A)*P(B)*not success of P(C) + P(A)*not success of P(B)*P(C) + not success of P(A)*P(B)*P(C)
= 1/4*1/2*(1-5/8) + 1/4*(1-1/2)*5/8 + 3/4*1/2*5/8
= 3/64 + 5/64 + 15/64
= 23/64
Answer (C)
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 23 Aug 2019, 08:28
the probabilities of success independently:
I -product: 1/4
II -product: 1/2
III -product: 5/8

What is the probability that exactly two of the launches succeed?
--> (successful) * (successful) * (NOT successful)=A
--> (successful) * (NOT successful) * (successful)=B
--> (NOT successful) * (successful) * (successful)=C

-> (\(\frac{1}{4})*(\frac{1}{2})*(1-\frac{5}{8}\)) =(\(\frac{1}{4})*(\frac{1}{2})*(\frac{3}{8}\))=\(\frac{3}{64}\)=A

-> (\(\frac{1}{4})*(1-\frac{1}{2})*(\frac{5}{8}\))=(\(\frac{1}{4})*(\frac{1}{2})*(\frac{5}{8}\))=\(\frac{5}{64}\)=B

-> (\(1-\frac{1}{4})(\frac{1}{2})\frac{(5}{8}\))=\(\frac{3}{4}*\frac{1}{2}*\frac{5}{8}\)=\(\frac{15}{64}\)=C

A+B+C=\(\frac{3}{64}+\frac{5}{64}+\frac{15}{64}\)=\(\frac{23}{64}\)

The answer is C.
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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New post 24 Aug 2019, 05:31
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Bunuel wrote:

Competition Mode Question



At a certain company, the probabilities of success of its three distinct new product launches are independently \(\frac{1}{4}\), \(\frac{1}{2}\), and \(\frac{5}{8}\), respectively. What is the probability that exactly two of the launches succeed?

A. \(\frac{7}{8}\)

B. \(\frac{1}{2}\)

C. \(\frac{23}{64}\)

D. \(\frac{5}{64}\)

E. \(\frac{3}{64}\)


We need to add the probabilities of the following scenarios:

Probability that the first two launches are successes and the last one is not:

1/4 x 1/2 x 3/8 = 3/64

and

Probability that the first and last launches are successes and the second one is not:

1/4 x 1/2 x 5/8 = 5/64

and

Probability that the last two launches are successes and the first one is not:

3/4 x 1/2 x 5/8 = 15/64

The probability oft exactly 2 successes is 3/64 + 5/64 + 15/64 = 23/64.

Answer: C
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Re: At a certain company, the probabilities of success of its three distin   [#permalink] 24 Aug 2019, 05:31
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