Question

A probability class has 30 students. As part of an assignment, each student tosses a coin 200 times and records the number of heads. Approximately what is the chance that no student gets exactly 100 heads?

Answer 1

Solution:

 The probability that a student gets exactly 100 heads out of 200

We have to combination(P&C) of 200 with 100 (200C100)/2^200  

= (200!)/(100! * 100! * 2^200) = P  

The probability that the student doesn't get exactly 100 heads out of 200 = 1-P  

The probability that all 30 students can't get exactly 100 heads out of 200 = (1-P)^30  

 [1- (200!)/(100! * 100! * 2^200)]^30

= 17.45% approximately.

This is the required solution.

Answer 2

Answer:

17.58% probability that no student gets exactly 100 heads

Step-by-step explanation:

We use the binomial probability distribution twice to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of a single student getting 100 heads.

The coin is tossed 200 times, so $n = 100$

For each toss, 50% probability of getting heads, so $p = 0.5$

Then

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 100) = C_{200,100}.(0.5)^{100}.(0.5)^{100} = 0.0563$

Approximately what is the chance that no student gets exactly 100 heads?

Each student has a 5.63% probability of getting exactly 100 heads, so $p = 0.0563$

30 students, so $n = 30$

We have to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{30,0}.(0.0563)^{0}.(0.9437)^{30} = 0.1758$

17.58% probability that no student gets exactly 100 heads