Question

The probability of occurrence of an event is 9/10. What is the probability that an event occur at least 4 times out of 5?

Answer 1

Answer:

The required probability is 0.32805

Step-by-step explanation:

The probability of occurrence of an event is 9/10. What is the probability that an event occur at least 4 times out of 5?

I have applied binomial distribution to solve this problem

The probability mass function of binomial distribution is

$\boxed{P(X=x)=n_C_x\:p^x\:q^{n-x}}$

Here, n=5

$\text{Probability of occurence }p=\frac{9}{10}$

$\text{Probability of non-occurence }q=\frac{1}{10}$

The required probability is

$=P(X=4)$

$=5_C_4\:(\frac{9}{10})^4\:(\frac{1}{10})^{5-4}$

$=5_C_1\:(\frac{9}{10})^4\:(\frac{1}{10})$

$=\frac{32805}{100000}$

$=0.32805$

Answer 2

Answer:

The probability that an event occur at least 4 times out of 5 is 0.91854.

Step-by-step explanation:

The probability of occurrence of an event is 9/10;

$P(E)=\frac{9}{10}$

Probability of not occurrence of that event

$P(\bar E) = 1 - \frac{9}{10} \\ \\ = \frac{1}{10}$

To find the probability of that event occur at least 4 times out of 5:

case 1: 4 success,1failure

(there are total 5 such cases,for ex.

SSSSF,SSSFS,SSFSS,SFSSS,FSSSS)

S: for success

F: failure

case 2:5 success

$= 5\big(P(E) \times \: P(E) \times P(E) \times P(E) \: \times P(\bar E)\big) + (P(E) \times \: P(E) \times P(E) \times P(E) \: \times P(E))$

$= 5\times \Big( \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} \times \frac{9}{10} \times \frac{1}{10}\Big) + \Big({ \frac{9}{10} }\Big)^{5}$

$= \Big({ \frac{9}{10} }\Big)^{4} \Big( \frac{5}{10} + \frac{9}{10} \Big)$

$= \Big({ \frac{9}{10} }\Big)^{4} \times \frac{14}{10}$

$= \frac{91854}{100000}$

$= \bf 0.91854$

Thus,

The probability that an event occur at least 4 times out of 5 is 0.91854.

Hope it helps you.