• Sep 7th 2010, 10:58 PM
cyt91
In a class of 30 students,10 of them come to school by bus,12 students walk to school and the others by car. 6 students are randomly selected. Find the probability that

(a) only 1 student come to school by car (ans:0.3548)

(b) at least 2 students come to school by bus (ans: 0.6736)

(c) equal number of students from each category is selected. (ans: 0.1401)

My attempt:

(a) Probability that a student comes to school by car is 8/30=4/15

X~B(6,4/15)

P(X=1) = 6C1 x (4/15) x (11/15)^5
= 0.3393

(b) Probability that a student comes to school by bus is 10/30 = 1/3

Y~B(6,1/3)

P(Y=>2) = 1- 6C1 x (1/3) x (2/3)^5 - 6C0 x (1/3)^0 x (2/3)^6
= 0.6488

(c) Probability that a student comes to school on foot is 12/30=2/5

F~B(6,2/5)

P(2F,2Y,2X) = 6C2 x (2/5)^2 x (3/5)^4 x 4C2 x (1/5)^2 x (2/5)^2 x 1
= 0.01194

What is the mistake in my approach?
• Sep 8th 2010, 12:17 AM
CaptainBlack
Quote:

Originally Posted by cyt91
In a class of 30 students,10 of them come to school by bus,12 students walk to school and the others by car. 6 students are randomly selected. Find the probability that

(a) only 1 student come to school by car (ans:0.3548)
My attempt:

(a) Probability that a student comes to school by car is 8/30=4/15

X~B(6,4/15)

No the number that come by car does not have a binomial distribtion.

The probability that the first student came by car is 8/30, then that the next did not come by car is 22/29, and then next 21/28, ...

So the required probbailty that the first came by car and the rest did not is:

(8x22x21xx20x19x18)/(30x29x29x27x26x27).

The total probabilty you require is six times this (to account for the cases where the child coming by car is the 1st, 2nd, ..., 6th).

CB
• Sep 8th 2010, 02:19 AM
cyt91
Ok. Thanks a lot. Is it correct to say that binomial distribution is used if the probability is constant for all trials?
• Sep 8th 2010, 02:28 AM
Unknown008
Well, the binomial distribution is used only when the probability is constant throughout all trials (Happy)

(with some other conditions as well)
• Sep 8th 2010, 03:15 AM
cyt91
Thanks. You guys have been helpful. (Happy)