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

Question

Competition Mode Question

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

A.  \frac{7}{8}

B.  \frac{1}{2}

C.  \frac{23}{64}

D.  \frac{5}{64}

E.  \frac{3}{64}

EncounterGMAT · Answer

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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Archit3110 · Answer

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

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

freedom128 · Answer

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)

MayankSingh · Answer

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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Kinshook · Answer

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

eakabuah · Answer

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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rishab0507 · Answer

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

Marcellodeog · Answer

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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Mohammadmo · Answer

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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joohwangie · Answer

\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}

kris19 · Answer

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)

lacktutor · Answer

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.

ScottTargetTestPrep · Answer

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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