The overall probability of a student failing an exam is 40%. Calculate the probability that in a group of 6 students, at least 4 passed the exam.

The average percentage of failure in a certain examination is 40. What is the probability that out of a group of 6 candidates, at least 4 passed in the examination?

Solution
Here the percentage of success, p = 0.60
And the percentage of failure q = 0.40
And number, n = 6
P (x ≥4) = ?
So
According to Binomial theory,
P (x ≥4) = P(x =4) + P (x =5)+ P (x =6)
= 6C4 〖.60〗^(4 ) 〖.40〗^(6-4) + 6C5 〖.60〗^(5 ) 〖.40〗^(6-5) + 6C6 〖.60〗^(6 ) 〖.40〗^(6-6)
= 0.31104 + 0.1866+ 0.046656
= 0.5436 (Answer)

This implies that either 4, 5 or all 6 will pass

prob = C(6,4) (.6)^4 (.4)^2 + C(6,5) (.6)^5 (.4) + C(6,6) .6^6

I get .54432

Thanks friend.

P(passed) = .6

P(failed) = .4//Compliment of P(passed)
N = 6
According to Binomial theory,
P (x ≥4) = P(x =4) + P (x =5)+ P (x =6)
= C(6,4) (.6)^4 (.4)^2 + C(6,5) (.6)^5 (.4) + C(6,6) .6^6
=0.544 Ans.

Thanku

To calculate the probability that at least 4 students passed the exam in a group of 6 students, we need to consider all possible scenarios:

1. All 6 students passed: This scenario has a probability of (0.6)^6 since each student has an independent probability of passing, which is 1 - 0.4 = 0.6. So, the probability of this scenario is 0.6^6.

2. 5 students passed: This scenario has a probability of (0.6)^5 * (0.4)^1, since we need 5 students to pass (probability of passing = 0.6) and 1 student to fail (probability of failing = 0.4). There are 6 ways to arrange 5 students passing and 1 student failing, so we multiply the individual probabilities by 6. Therefore, the probability of this scenario is 6 * (0.6)^5 * (0.4)^1.

3. 4 students passed: This scenario has a probability of (0.6)^4 * (0.4)^2, since we need 4 students to pass (probability of passing = 0.6) and 2 students to fail (probability of failing = 0.4). There are 15 ways to arrange 4 students passing and 2 students failing, so we multiply the individual probabilities by 15. Therefore, the probability of this scenario is 15 * (0.6)^4 * (0.4)^2.

To calculate the probability that at least 4 students passed, we need to sum up the probabilities of the above three scenarios:

Probability(at least 4 students passed) = Probability(all 6 passed) + Probability(5 passed) + Probability(4 passed)

Therefore,

Probability(at least 4 students passed) = 0.6^6 + 6 * (0.6)^5 * (0.4)^1 + 15 * (0.6)^4 * (0.4)^2.

Evaluating this expression will give you the desired probability.

4 5 or 6

p(passed) = .6
p(failed) = .4

Binomial distribution
for example for 4 of 6 passed
C(6,4) .6^4 .4^2
C(6,4) = 6!/[ 4!(2!)] = 6*5/2 = 15
15 * .6^4*.4^2 = .311

do that also for 6,5
and for 6,6
and add those three results