GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 16 Sep 2019, 09:34

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 57993
At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

22 Aug 2019, 07:09
1
2
00:00

Difficulty:

35% (medium)

Question Stats:

80% (01:49) correct 20% (02:02) wrong based on 75 sessions

### HideShow timer Statistics

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. $$\frac{3}{64}$$

_________________
Senior Manager
Status: Whatever it takes!
Joined: 10 Oct 2018
Posts: 328
GPA: 4
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

22 Aug 2019, 07:17
2
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.

Posted from my mobile device
Attachments

IMG_20190822_194331.jpg [ 1.26 MiB | Viewed 968 times ]

GMAT Club Legend
Joined: 18 Aug 2017
Posts: 4729
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

22 Aug 2019, 07:23
1
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
_________________
If you liked my solution then please give Kudos. Kudos encourage active discussions.
Senior Manager
Joined: 30 Sep 2017
Posts: 336
Concentration: Technology, Entrepreneurship
GMAT 1: 720 Q49 V40
GPA: 3.8
WE: Engineering (Real Estate)
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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

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.
Manager
Joined: 08 Jan 2018
Posts: 145
Location: India
Concentration: Operations, General Management
WE: Project Management (Manufacturing)
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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

Posted from my mobile device
VP
Joined: 03 Jun 2019
Posts: 1498
Location: India
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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
_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."

Please provide kudos if you like my post. Kudos encourage active discussions.

My GMAT Resources: -

Efficient Learning
All you need to know about GMAT quant

Tele: +91-11-40396815
Mobile : +91-9910661622
E-mail : kinshook.chaturvedi@gmail.com
Senior Manager
Joined: 18 May 2019
Posts: 250
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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

Posted from my mobile device
Manager
Joined: 12 Mar 2019
Posts: 154
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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

Intern
Joined: 21 Aug 2019
Posts: 1
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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

Posted from my mobile device
Senior Manager
Joined: 29 Jun 2019
Posts: 300
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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

Posted from my mobile device
_________________
Always waiting
Intern
Joined: 17 Jan 2019
Posts: 48
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

23 Aug 2019, 00:45
$$\frac{1}{4}x\frac{1}{2}x\frac{3}{8}$$
+
$$\frac{1}{4}x\frac{1}{2}x\frac{5}{8}$$
+
$$\frac{3}{4}x\frac{1}{2}x\frac{5}{8}$$
=$$\frac{3}{64}+\frac{5}{64}+\frac{15}{64}$$
=$$\frac{23}{64}$$
_________________
kudos for kudos? :P
Manager
Joined: 24 Sep 2014
Posts: 50
Concentration: General Management, Technology
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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
Manager
Joined: 25 Jul 2018
Posts: 207
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

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=$$\frac{3}{64}+\frac{5}{64}+\frac{15}{64}$$=$$\frac{23}{64}$$

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 7706
Location: United States (CA)
Re: At a certain company, the probabilities of success of its three distin  [#permalink]

### Show Tags

24 Aug 2019, 05:31
1
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.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: At a certain company, the probabilities of success of its three distin   [#permalink] 24 Aug 2019, 05:31
Display posts from previous: Sort by