Back to Exercise: Develop a theoretical probability distribution

Exercises: Develop a Theoretical Probability Distribution and Find Its Expected Value

Work through each section in order. Show your work where indicated. When you build a probability,
separate "the probability of one specific sequence" from "the number of arrangements" that produce
that result. Keep fractions over a common denominator so your sum-to-1 check is easy to read.

Grade 11·24 problems·~45 min·Common Core Math - HS Statistics and Probability·group·hss-md-a-3
Printable layout
A

Recall / Warm-Up

1.

On a multiple-choice test each question has four choices and a guesser is correct with
probability 14\frac{1}{4}. The questions are independent. What is the probability of getting
the first two specific questions both correct?

A.

14+14=12\frac{1}{4} + \frac{1}{4} = \frac{1}{2}

B.

14×14=116\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

C.

14\frac{1}{4}

D.

24=12\frac{2}{4} = \frac{1}{2}

2.

A student guesses on every question of a five-question test where each question has four
equally likely choices. Let XX be the number of questions answered correctly. Which statement
correctly describes the model?

A.

P(correct)=14P(\text{correct}) = \frac{1}{4}, P(wrong)=34P(\text{wrong}) = \frac{3}{4}, and the questions are independent.

B.

P(correct)=12P(\text{correct}) = \frac{1}{2} because each answer is either right or wrong.

C.

P(correct)=14P(\text{correct}) = \frac{1}{4}, but the questions are dependent on one another.

D.

P(correct)=34P(\text{correct}) = \frac{3}{4} because there are three wrong choices.

3.

How many different ways are there to choose which 3 of the 5 questions are the correct ones?
That is, evaluate (53)\binom{5}{3}.

B

Fluency Practice

1.

For the five-question guessing test (P(correct)=14P(\text{correct}) = \frac{1}{4}), which expression
correctly gives P(X=2)P(X = 2), the probability of exactly 2 correct?

A.

(52)(14)2(34)3\binom{5}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3

B.

(14)2(34)3\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3

C.

16\frac{1}{6}, since XX has six possible values 0,1,2,3,4,50,1,2,3,4,5

D.

(52)(14)2\binom{5}{2}\left(\frac{1}{4}\right)^2

2.

For the five-question guessing test, compute P(X=2)=(52)(14)2(34)3P(X = 2) = \binom{5}{2}\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^3.
Write your answer as a fraction over a denominator of 10241024.

3.

For the five-question guessing test, compute P(X=3)=(53)(14)3(34)2P(X = 3) = \binom{5}{3}\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^2.
Write your answer as a fraction over a denominator of 10241024.

4.

The full developed distribution for the five-question guessing test is shown below.

kk012345
P(X=k)P(X=k)2431024\frac{243}{1024}4051024\frac{405}{1024}2701024\frac{270}{1024}901024\frac{90}{1024}151024\frac{15}{1024}11024\frac{1}{1024}

Add the six numerators to verify the distribution is valid. What is the sum of the numerators?

5.

Using the distribution table above, compute the expected number correct
E(X)=kP(X=k)E(X) = \sum k \cdot P(X=k) for the five-question guessing test. Give a decimal.

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