1. ## Spot my mistake please...

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?

2. 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

3. Ok. Thanks a lot. Is it correct to say that binomial distribution is used if the probability is constant for all trials?

4. Well, the binomial distribution is used only when the probability is constant throughout all trials

(with some other conditions as well)

5. Thanks. You guys have been helpful.