Test Questions Probability

[WalkingMullet]

Can someone please help me with this problem??

A test has 5 true and false quesions and 5 multiple choice questions. Each multiple choice question has 4 possible answers. If Maria guesses on all 10 questions, what is the probablity that she will get exactly 8 right answers?

There's 3 choices: A. 189/2048 B. 175/ 32,768 C. 243/1024

I think you start it something like (1/2)(1/2)(1/2)(1/2)(1/2)(1/4)(1/4)(1/4)(1/4)(1/4), but I'm not really sure where the exactly 8 comes into play.

If anyone could help explain this to me I would really appreciate it!

 

[soroban]

Hello, WalkingMullet!

This is trickier than you think . . .

A test has 5 true-false quesions and 5 multiple-choice questions.
Each multiple choice question has 4 possible answers.
If Maria guesses on all 10 questions, what is the probablity
that she will get exactly 8 right answers?

There are three ways she can get 8 right answers . . .

- . . (1) Get 5 T/F questions correct and 3 M/C questions correct
or: .(2) Get 4 T/F questions correct and 4 M/C questions correct.
or: .(3) Get 3 T/F questions correct and 5 M/C questions correct.

Case 1: \(\displaystyle \(\text{5 T/F and 3 M/C}) \;=\;{5\choose5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^0\,\cdot\,{5\choose3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2 \;=\;\frac{90}{2^{15}}\)

Case 2: \(\displaystyle \(\text{4 T/F and 4 M/C}) \;=\;{5\choose4}\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^1 \,\cdot\,{5\choose4}\left(\frac{1}{4}\right)^4\left(\frac{3}{4}\right)^1\;=\;\frac{75}{2^{^{15}}}\)

Case 3: \(\displaystyle \(\text{3 T/F and 5 M/C}) \;=\;{5\choose3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^2\,\cdot\,{5\choose5}\left(\frac{1}{4}\right)^5\left(\frac{3}{4}\right)^0 \;=\;\frac{10}{2^{^{15}}}\)


Therefore: \(\displaystyle \L\(\text{8 right}) \;=\;\frac{90}{2^{15}}\,+\,\frac{75}{2^{15}}\,+\,\frac{10}{2^{^{15}}} \;=\;\frac{175}{32,768}\)

 

[pka]

Note it could be 3 true/false and 5 multiple choice, 4 true/false and 4 multiple choice, or 5 true/false and 3 multiple choice.
\(\displaystyle \L\begin{array}{ccl}
T &\| & M &\| & {P(X = 8)} \\
\hline
3 &\| & 5 &\| & {\left( {\begin{array}{c}
5 \\
3 \\
\end{array}} \right)\left( {.5} \right)^5 \left( {\begin{array}{c}
5 \\
5 \\
\end{array}} \right)\left( {.25} \right)^5 } \\
4 &\| & 4 &\| & {\left( {\begin{array}{c}
5 \\
4 \\
\end{array}} \right)\left( {.5} \right)^5 \left( {\begin{array}{c}
5 \\
4 \\
\end{array}} \right)\left( {.25} \right)^4 \left( {.75} \right)} \\
5 &\| & 3 &\| & {\left( {\begin{array}{c}
5 \\
5 \\
\end{array}} \right)\left( {.5} \right)^5 \left( {\begin{array}{c}
5 \\
3 \\
\end{array}} \right)\left( {.25} \right)^3 \left( {.75} \right)^2 } \\
\end{array}\).

Now addup the probabilities.

 

[WalkingMullet]

Thank you very much!! :wink: I really appreciate it!