Study the binomial distribution table. Notice that the proba

Step 1
a) Given that ${\displaystyle n=5}$
Consider P(3 successes), if ${\displaystyle p=0.30}$
$$P(3successes)=(\begin{array}{c}5\\ 3\end{array}){0.3}^3(1-0.3{)}^{5-3}=0.1323$$
Consider P(2 or fewer successes), where ${\displaystyle p=0.7}$
$P(2\text{or fewer successes})=P(0success)+P(1success)+P(2success)=(\begin{array}{c}5\\ 0\end{array}){0.7}^0(1-0.7{)}^{5-0}+(\begin{array}{c}5\\ 1\end{array}){0.7}^1(1-0.7{)}^{5-1}+(\begin{array}{c}5\\ 2\end{array}){0.7}^2(1-0.7{)}^{5-2}=0.1631$
Here, the value of P(3 successes) is less than P(2 or fewer successes).
P(3 successes), where ${\displaystyle p=0.30}$, is smaller than P(2 or fewer successes), where ${\displaystyle p=0.70}$.
Step 2
b) Given that ${\displaystyle n=5}$
Consider P(3 or more successes), if ${\displaystyle p=0.30}$
$P(ormoresuccesses)=(\begin{array}{c}5\\ 3\end{array}){0.3}^3(1-0.3{)}^{5-3}+(\begin{array}{c}5\\ 4\end{array}){0.3}^4(1-0.3{)}^{5-4}+(\begin{array}{c}5\\ 5\end{array}){0.3}^5(1-0.3{)}^{5-5}=0.1631$
Consider P(2 or fewer successes), where ${\displaystyle p=0.7}$
${\displaystyle P\left(\text{2 or fewer successes}\right)=P\left(0success\right)+P\left(1success\right)+P\left(2success\right)}$
$=(\begin{array}{c}5\\ 0\end{array}){0.7}^0(1-0.7{)}^{5-0}+(\begin{array}{c}5\\ 1\end{array}){0.7}^1(1-0.7{)}^{5-1}+(\begin{array}{c}5\\ 2\end{array}){0.7}^2(1-0.7{)}^{5-2}=0.1631$
Here, the value of P(3 or more successes) is same as P(2 or fewer successes).
P(3 or more successes), where ${\displaystyle p=0.30}$, is same as P(2 or fewer successes), where ${\displaystyle p=0.70}$.
Step 3
c) P(4 successes), where ${\displaystyle p=0.30}$.
$$P(4successes)=(\begin{array}{c}5\\ 4\end{array}){0.3}^4(1-0.3{)}^{5-4}=0.028$$

Step 1
a) From table:
${\displaystyle P\left(3successes\right)=0.1323}$ where ${\displaystyle p=0.30}$ and ${\displaystyle n=5}$
${\displaystyle P\left(2successes\right)=0.1323}$, where ${\displaystyle p=0.70}$ and ${\displaystyle n=5}$
They are the same.
b) ${\displaystyle P\left(3\text{or more successes}\right)=0.1323+0.0284+0.0024=0.1631,}$ where ${\displaystyle p=0.30}$ and ${\displaystyle n=5}$
${\displaystyle P\left(3\text{or fewer successes}\right)=0.1323+0.0284+0.0024=0.1631,}$ where ${\displaystyle p=0.70}$ and ${\displaystyle n=5}$
They are the same.
c) ${\displaystyle P\left(4successes\right)=0.0284}$, where ${\displaystyle p=0.30}$ and ${\displaystyle n=5}$
${\displaystyle P\left(1successes\right)=0.0284}$, where ${\displaystyle p=0.70}$ and ${\displaystyle n=5}$
So, ${\displaystyle r=1}$
d) the column headed by ${\displaystyle p=0.80}$

Step 1 Compare P(3 successes), where p = 0.30, with P(2 successes), where p = 0.70. P(3 successes)=0.1323P( 2 successes) =0.1323 P(3 successes), where p = 0.30, is larger. Answer: They are the same. P(3 successes), where p = 0.30, is smaller. NSK b) Compare P(3 or more successes), where p = 0.30, with P(2 or fewer successes), where p = 0.70. P( 3 or more)=0.163 P( 2 or fewer) =0.163 P(3 or more successes), where p = 0.30, is larger. Answer: They are the same. P(3 or more successes), where p = 0.30, is smaller. c) Find the value of P(4 successes), where p = 0.30. (Round your answer to three decimal places.) P(4 successes) = 0.028 For what value of r is P(r successes) the same using p = 0.70? Answer: r = 1 d) What column is symmetrical with the one headed by p = 0.20? Answer: the column headed by p = 0.80 the column headed by p=0.40 the column headed by p=0.70 the column headed by p = 0.50