Answer:
[tex]P(x \le 3) = 0.2352[/tex]
Step-by-step explanation:
Given
[tex]p = 0.35[/tex] --- male committing a crime in an episode
[tex]n = 5[/tex] -- Number of episodes
Required
Determine the probability of male committing a crime at least 3 times
This question illustrates binomial distribution and will be solved using;
[tex]P(x) = ^nC_x * p^x * (1 - p)^{n-x[/tex]
So, the required probability is represented as:
[tex]P(x \ge 3)[/tex]
And will be calculated using:
[tex]P(x \ge 3) = P(x = 3) + P(x = 4) + P(x = 5)[/tex]
[tex]P(x = 3) = ^5C_3 * (0.35)^3 * (1 - 0.35)^{5-3}[/tex]
[tex]P(x = 3) = ^5C_3 * (0.35)^3 * (1 - 0.35)^2[/tex]
[tex]P(x = 3) = 10 * (0.35)^3 * (0.65)^2[/tex]
[tex]P(x = 3) = 0.1811[/tex]
[tex]P(x = 4) = ^5C_4 * (0.35)^4 * (1 - 0.35)^{5-4}[/tex]
[tex]P(x = 4) = 5 * (0.35)^4 * (1 - 0.35)^1[/tex]
[tex]P(x = 4) = 5 * (0.35)^4 * (0.65)[/tex]
[tex]P(x = 4) = 0.0488[/tex]
[tex]P(x = 5) = ^5C_5 * (0.35)^5 * (1 - 0.35)^{5-5}[/tex]
[tex]P(x = 5) = 1 * (0.35)^5 * (1 - 0.35)^0[/tex]
[tex]P(x = 5) = 1 * (0.35)^5 * (0.65)^0[/tex]
[tex]P(x = 5) = 1 * (0.35)^5 * 1[/tex]
[tex]P(x = 5) = 0.0053[/tex]
So:
[tex]P(x \ge 3) = P(x = 3) + P(x = 4) + P(x = 5)[/tex]
[tex]P(x \le 3) = 0.1811 + 0.0488 + 0.0053[/tex]
[tex]P(x \le 3) = 0.2352[/tex]
