Question

Lost-time accidents occur in a company at a mean rate of 0.3 per day. What is the probability that the number of lost-time accidents occurring over a period of 9 days will be no more than 3? Round your answer to four decimal places.

Answer 1

Answer:

0.7295

Step-by-step explanation:

Given that :

Mean rate of lost time accident = 0.3 per day

probability that the number of lost-time accidents occurring over a period of 9 days will be no more than 3?

n = 9

Using binomial distribution formula :

P(x ≤3)

Probability of success (p) = 0.3

1 - p = 1- 0.3 = 0.7

nCx * p^x * (1 - p)^(n-x)

Using the binomial probability distribution calculator to save computation time :

P(x ≤3) = P(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

P(x ≤3) = 0.0403 + 0.1556 + 0.2668 + 0.2668

P(x ≤3) = 0.7295

Answer 2

Probabilities are used to determine the chances of events.

The probability that accidents occur no more than 3 days in 9 days is 0.7296

The given parameters are:

$p = 0.3$ --- the probability that an accident occurs

The probability that an accident does not occur (q) is then calculated using the following complement rule

$q= 1 - p$

$q= 1 - 0.3$

$q= 0.7$

The probability that accidents occur no more than 3 days in 9 days is

$P(x \le 3) = P(0) + P(1) + P(2) + P(3)$

Each probability is calculated using:

$P = ^nC_rp^rq^{n-r}$

So, we have:

$P(x \le 3) = ^9C_0(0.3)^0(0.7)^9 +^9C_1(0.3)^1(0.7)^8 + ^9C_2(0.3)^2(0.7)^7 + ^9C_3(0.3)^3(0.7)^6$

This gives

$P(x \le 3) = 0.0403 + 0.1556 + 0.2668 + 0.2669$

$P(x \le 3) = 0.7296$

Hence, the probability that accidents occur no more than 3 days in 9 days is 0.7296

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