Question

let the probability of a patient recovering from a psychological disorder after a treatment is 0.75. suppose there are 4 patients. X denotes number of patients recovered. find the probability of p(X=3)

Answer 1

Given :  the probability of a patient recovering from a psychological disorder after a treatment is 0.75.  

there are 4 patients.

X denotes number of patients recovered.

To find : the probability of p(X=3)

Solution:

p = 0.75 = 3/4

q  = 1 - p = 1 - 3/4 = 1/4

n = 4

X = 3

P(x) = ⁿCₓpˣqⁿ⁻ˣ

P(3) =  ⁴C₃ ( 3/4)³ (1/4)¹

=27/64

=  0.421875

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Answer 2

Answer:

The probability of $P(X=3)$ is $0.421875$.

Step-by-step explanation:

Formula for binomial distribution probability:

$P(X=x)=n_C_x \ a^xb^{n-x}$, where

$P$ = binomial probability,

$x$ = number of times for an outcome within  trails,

$n_C_x$ = number of combinations

$a$ = Probability of $success$

$b$ = Probability of $failure$

$n$ = number of trials

According to the question,

X denotes the number of patients recovered.

After a treatment, probability of recovering a patient from a psychological disorder = 0.75

i.e., probability of success, $a=0.75$

So, the probability of failure, $b=1-0.75$

                                                  $=0.25$

It is given that number of patients are 4.

i.e., $n=4$

To find: $P(X=3)$

Using binomial distribution,

$P(X=3)=n_C_3 \ a^3b^{n-3}$

On substituting the values, we get

$P(X=3)=4_C_3 \ (0.75)^3(0.25)^{4-3}$

               $=\frac{4!}{3!(4-3)!} \ (0.75)^3(0.25)$

               $=\frac{4\times 3!}{3!1!} \ (0.75)^3(0.25)$

Further, simplify as follows:

$P(X=3)=\frac{4}{1} (0.421875)(0.25)$

                $=0.421875$

Therefore, the probability of $P(X=3)$ is $0.421875$.

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