Question

hi, can you help me answer this question, please, thank you:)

Answer 1

The question requires that we use the binomial probability distribution

To do this, we will use the binomial formula

$^nC_r\times P^r\times q^{n-r}$

Where

$\begin{gathered} P=\text{ probability of success} \\ q=\text{ probabilty of failure} \\ n=\text{ number of times} \\ \end{gathered}$

In our case

$P=76\text{ \%=}\frac{76}{100}=0.76$$\begin{gathered} q=1-p=1-0.76=0.24 \\ q=0.24 \end{gathered}$

$n=12$

For the first question,

We are told to find the probability that exactly 6 jurors are Asian American

In this case,

$\begin{gathered} r=6 \\ \end{gathered}$

So applying the formula

$\begin{gathered} (P=6)=>^{12}C_6\times0.76^6\times0.24^{12-6}=^{12}C_6\times0.76^6\times0.24^6 \\ \\ (P=6)\Rightarrow924\times0.1927\times1.911\times10^{-4}=0.0340 \\ \\ \end{gathered}$

Thus, the probability that exactly 6 jurors are Asian American =0.0340

For the second question

We are told to find the probability of 6 or fewer than 6 are Asian Americans

$\begin{gathered} \text{When r=0} \\ (P=0)\Rightarrow^{12}C_0\times0.76^0\times0.24^{12}=^{12}C_0\times0.76^0\times0.24^{12} \\ (P=0)=1\times1\times3.65^{}\times10^{-8} \\ P\text{ is appro}\xi mately\text{ 0} \end{gathered}$

We will repeat this for when r=1, 2, 3, 4, 5 and 6

And then we will sum the values up

$\begin{gathered} \text{When r=1} \\ P=\text{appro}\xi\text{mately 0} \end{gathered}$

$\begin{gathered} when\text{ r=2} \\ P\approx0.0002 \end{gathered}$

$\begin{gathered} \text{when r=3} \\ P\approx0.00026 \end{gathered}$$\begin{gathered} \text{when r=4} \\ P\approx0.00182 \end{gathered}$

$\begin{gathered} \text{when r=5} \\ P=0.00921 \end{gathered}$

$\begin{gathered} \text{when r=6} \\ P=0.03403 \end{gathered}$

Then

The probability of 6 or fewer than 6 are Asian Americans = P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)

Thus,

The probability of 6 or fewer than 6 are Asian Americans = 0+0+0.00002+0.00026+0.00182+0.00921+0.03403=0.04534

The probability of 6 or fewer than 6 is Asian Americans=0.04534

For the third part

The probability of more than 10 will be

The sum of the probabilities when r = 11 and r =12

$\begin{gathered} \text{when r=11} \\ ^{12}C_{11}\times0.76^{11}\times0.24^{12-11}=^{12}C_{11}\times0.76^{11}\times0.24^1=0.14072 \end{gathered}$

$\begin{gathered} \text{when r=12} \\ ^{12}C_{12}\times0.76^{12}\times0.24^{12-12}=^{12}C_{12}\times0.76^{12}\times0.24^0=0.03713 \end{gathered}$

Thus, the total probability = P(11) + P(12)= 0.14072 +0.03713 = 0.17785

Hence,

The probability of more than 10 Asian-Americans will be = 0.17785