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 Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

In an insurance company, each policy has a paper record and [#permalink]
17 Aug 2003, 23:22

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?

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?

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?

Answer is 0.94

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

Re: Probability Q2 [#permalink]
18 Aug 2003, 06:34

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)

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?