Given:
P(4 sucessses) n=15 P=0.04
P(2 successors) n=12. P=0.2
P( at most 3 sucesssors) n= 20 p=0.5
Required:
To calculate above equation
Explanation:
a)-P(4 sucessses) n=15 P=0.04
a=no of trials
p=probability of success
[tex]\begin{gathered} P(4\text{ successes\rparen=P\lparen x=4\rparen} \\ \\ USING: \\ \\ ^nc_x\times p\text{ x }\times(1-p)(n-x) \\ \\ ^{15}c_4\times0.4\text{ 4}\times(1-0.4)\div(15-4) \\ \\ =0.126775 \end{gathered}[/tex]b)-
n=12,p=0.2, find P(2 failures),
P(2 failures)=P(12-2)=p(10 success)
using the same formula after calculation we get
=0.000004325376
c)-n=20, p=0.05 , find P(at least 3 successes)
P(x>3)=p(3)+p(4)+p(5)........p(20)
To avoid the complicated calculations we can use the online bionomial probability distribution calculator we get 0.05748
Required answer:
a)- 0.126775
b)-=0.000004325376
c)-0.05748