The value of the probability P(x > 2) is 0.8369
How to evaluate the probability?
The given parameters are:
n = 5
p =0.7
The probability is calculated as:
[tex]P(x) = ^nC_x *p^x * (1 - p)^x[/tex]
Using the complement rule, we have:
P(x > 2) = 1 - P(0) - P(1) - P(2)
Where:
[tex]P(0) = ^5C_0 *0.7^0 * (1 - 0.7)^5[/tex]
P(0) = 1 *1 * (1 - 0.7)^5 = 0.00243
[tex]P(1) = ^5C_1 *0.7^1 * (1 - 0.7)^4[/tex]
P(1) = 5 *0.7^1 * (1 - 0.7)^4 = 0.02835
[tex]P(2) = ^5C_2 *0.7^2 * (1 - 0.7)^3[/tex]
P(2) = 10 *0.7^2 * (1 - 0.7)^3 = 0.1323
Recall that:
P(x > 2) = 1 - P(0) - P(1) - P(2)
So, we have:
P(x > 2) = 1 - 0.00243 - 0.02835 - 0.1323
Evaluate
P(x > 2) = 0.83692
Approximate
P(x > 2) = 0.8369
Hence, the value of the probability P(x > 2) is 0.8369
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