Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Aug 2016, 07:29

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In an insurance company, each policy has a paper record and

Author Message
Manager
Joined: 30 May 2003
Posts: 92
Location: Toronto
Followers: 1

Kudos [?]: 2 [0], given: 0

In an insurance company, each policy has a paper record and [#permalink]

### Show Tags

18 Aug 2003, 00:22
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

23. In an insurance company, each policy has a paper record and an
electric record. For those policies having incorrect paper record,
60% also having incorrect electric record; For those policies having incorrect electric record, 75% also having incorrect paper record. 3% of all policies have both incorrect paper and incorrect electric records. If we randomly pick out one policy, what's the probability that the one having both correct paper and correct electric records?
Manager
Joined: 30 May 2003
Posts: 92
Location: Toronto
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

18 Aug 2003, 00:26
A's breakfast has 2 chiken sandwish and 4 cheese sandwish;
B's breakfast has 3 chiken sadwish and 5 cheese sandwish. Choosing 1 from each meal randomly, what is the probability of both are chicken?
Manager
Joined: 30 May 2003
Posts: 92
Location: Toronto
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

18 Aug 2003, 00:57
are we not considering the probability of choosing 1 out of the 2 dishes i.e. 1/2*1/8=1/16????
Manager
Joined: 28 Feb 2003
Posts: 147
Location: Kiev
Followers: 1

Kudos [?]: 6 [0], given: 0

### Show Tags

18 Aug 2003, 02:01
agree with aps_can

_________________

Too much is not enough...

Intern
Joined: 18 Jul 2003
Posts: 10
Location: Mumbai
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

18 Aug 2003, 02:15
Answer to 1st one = 0.88

Answer to the sandwich question = 17/48
Manager
Joined: 11 Jun 2003
Posts: 72
Location: Moscow, Russia
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

18 Aug 2003, 07:06
3%/60% + 3%/75% - 3% = 6%
100% - 6% = 94%
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2

Kudos [?]: 2 [0], given: 0

### Show Tags

18 Aug 2003, 07:31
prashant wrote:
aps_can wrote:
23. In an insurance company, each policy has a paper record and an
electric record. For those policies having incorrect paper record,
60% also having incorrect electric record; For those policies having incorrect electric record, 75% also having incorrect paper record. 3% of all policies have both incorrect paper and incorrect electric records. If we randomly pick out one policy, what's the probability that the one having both correct paper and correct electric records?

Here's the explanation:

Let the total no. of policies be T
Let no. of incorrect paper be x
Let no. of incorrect electric be y

Now 0.03*T = 0.6*x = 0.75*y

The required probability is [T-(x+y-0.03*T)]/T

Substituting and solving, we get the required prob. as 0.94
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2

Kudos [?]: 2 [0], given: 0

### Show Tags

18 Aug 2003, 07:34
prashant wrote:
aps_can wrote:
A's breakfast has 2 chiken sandwish and 4 cheese sandwish;
B's breakfast has 3 chiken sadwish and 5 cheese sandwish. Choosing 1 from each meal randomly, what is the probability of both are chicken?

1/8

P(both chicken) = P(chicken from meal 1)*P(chicken from meal 2)

= 2/6*3/8 = 1/8
Manager
Joined: 30 May 2003
Posts: 92
Location: Toronto
Followers: 1

Kudos [?]: 2 [0], given: 0

### Show Tags

18 Aug 2003, 08:32
Thanks for the solving Q1 & Q2

In how many ways 30 couples(total=30 men and 30 women...with respective man sitting on the right of woman) and 30 HOD(head of the department) be seated around a round table if each of the couple should sit between each of the HOD?
Probability Q3   [#permalink] 18 Aug 2003, 08:32
Display posts from previous: Sort by