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Math Expert V
Joined: 02 Sep 2009
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At a certain company, the probabilities of success of its three distin  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 80% (01:50) correct 20% (02:06) wrong based on 80 sessions

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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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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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Attachments IMG_20190822_194331.jpg [ 1.26 MiB | Viewed 1047 times ]

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

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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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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

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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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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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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Senior Manager  G
Joined: 18 May 2019
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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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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Joined: 12 Mar 2019
Posts: 160
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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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

Intern  Joined: 21 Aug 2019
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Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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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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Senior Manager  G
Joined: 29 Jun 2019
Posts: 408
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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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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Posts: 87
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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$$\frac{1}{4}x\frac{1}{2}x\frac{3}{8}$$
+
$$\frac{1}{4}x\frac{1}{2}x\frac{5}{8}$$
+
$$\frac{3}{4}x\frac{1}{2}x\frac{5}{8}$$
=$$\frac{3}{64}+\frac{5}{64}+\frac{15}{64}$$
=$$\frac{23}{64}$$
Manager  S
Joined: 24 Sep 2014
Posts: 51
Concentration: General Management, Technology
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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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
Senior Manager  G
Joined: 25 Jul 2018
Posts: 258
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

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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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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.

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