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probability
by any 203 - Thursday, 23 September 2010, 02:37 PM
 
kindly  explain the answer  for following  question

 A certain roller coaster has 3 cars ,and a passenger is equally likely to ride in any 1 of the 3 cars each time the passenger rides the roller coaster . If a certain passenger is to ride the roller coaster three times , what is the  probability that the passenger  will ride  in each of the three cars ?

Re: probability
by TG Team - Thursday, 23 September 2010, 05:21 PM
  It is 3!(1/3)3 = 2/9. smile
Re: probability
by sidhartha nayak - Thursday, 23 September 2010, 11:21 PM
  Another Question:..Plz explain how to approach the following question..

Two friends agree to meet at a park with the
following conditions. each will reach the park
between 4:00 pm and 5:00 pn and will see if
the other has already arrived. If not, they will
wait for 10 minutes or the end of the hour
which ever is earlier and leave. what is the
probability that the two will not meet?
Re: probability
by TG Team - Friday, 24 September 2010, 01:37 PM
 

Hi Sidharthasmile

One approach is using geometric probability excellently explained by TG in his lesson - Probability.

Another basic one is - make two cases - (i) One who is arriving earlier reaches within first 50 minutes and the other one arrives in next adjacent 10 minutes. So required probability in this case comes out to be = 2*(5/6)(1/6) = 10/36.

(ii) Both friends are arriving in last 10 minutes. So probility for this case = (1/6)(1/6) = 1/36.

Hence total probability of meeting of two friends with given conditions becomes = 10/36 + 1/36 = 11/36. smile

Re: probability
by any 203 - Monday, 27 September 2010, 01:14 PM
  Hi Kamal ,

thanks for prompt reply ,
i am still unable to grasp the logic for roller coaster question . If possible kindly elaborate .
Re: probability
by TG Team - Monday, 27 September 2010, 02:08 PM
 

Total cases: 33 = 27 as the passenger can ride on any car on all the three instances.

Favorable cases: 3! = 6 as the passenger need to ride on all the three cars on all the three instances one at a time. The order of riding can be arranged in 3! ways.

So the rewuired probability = 6/27 = 2/9.smile

Re: probability
by Puneet Goyal - Wednesday, 29 September 2010, 12:09 AM
  The other way of thinking about the roller coaster problem is:

For first time, the guy can sit in any cart, thus the probability is 3*1/3 .
For second time, he can sit in any of the other two carts, thus 2*1/3.
And third time he has to sit in the last cart, 1*13.

Thus total probability is 3*2*1 * 1/3 * 1/3 1/3 = 2/9.
 
Re: probability
by Puneet Goyal - Wednesday, 29 September 2010, 12:12 AM
  One DS ques on Probability:

If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p >1/2 ?

(1)   More than 1/2of the 10 employees are women.

(2)   The probability that both representatives selected will be men is less than 1/10.