Answer with Step-by-step explanation:
For a given Binomial event with probability of success 'p' the probabolity of 'r' success in 'n' trails is given by
[tex]P(E)=\frac{n!}{(n-r)!\cdot r!}p^r\cdot (1-p)^{n-r}[/tex]
Thus for given 4 trails (n = 4) we have
Part 1)
Probability of 1 success is
[tex]P(E_1)=\frac{4!}{(4-1)!\cdot 1!}\cdot 0.7^1\cdot (1-0.7)^{4-1}\\\\\therefore P(E_1)=0.076[/tex]
Part 2)
Probability of 2 successes
[tex]P(E_2)=\frac{4!}{(4-2)!\cdot 2!}\cdot 0.7^2\cdot (1-0.7)^{4-2}\\\\\therefore P(E_2)=0.265[/tex]
Part 3)
Probability of at least 1 success
[tex]P(E)=1-(1-p)^n\\\\\therefore P(E_3)=1-(1-0.7)^{4}=0.992[/tex]