Question

1. Assume that random guesses are made for 8 multiple-choice questions on a test with 2 choices for each​ question, so that there are n=8

trials, each with probability of success​ (correct) given by p equals=0.50. Find the probability of no correct answers

2. A survey showed that 78​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If 11 adults are randomly​ selected, find the probability that at least at least 10 of them need correction for their eyesight. Is 10 a significantly high number of adults requiring eyesight​ correction?

A. The probability that at least 10 of the 11 adults require eyesight correction is what?

B. Is 10 a significantly high number of adults requiring eyesight​ correction? Note that a small probability is one that is less than 0.05.

a. Yes​, because the probability of this occurring is small.

b.No​, because the probability of this occurring is small.

C. Yes​, because the probability of this occurring is not is not small.

D. No​, because the probability of this occurring is not is not small.

3. In a​ state's Pick 3 lottery​ game, you pay $1.14 to select a sequence of three digits​ (from 0 to​ 9), such as 311.

If you select the same sequence of three digits that are​ drawn, you win and collect $287.33.

Complete parts​ (a) through​ (e).

a. How many different selections are​ possible?

b. What is the probability of​ winning?

​(Type an integer or a​ decimal.)

c. If you​ win, what is your net​ profit?

​$

​(Type an integer or a​ decimal.)

d. Find the expected value.

​$

​(Round to the nearest hundredth as​ needed.)

e. If you bet $1.14 in a certain​ state's Pick 4​ game, the expected value is negative −$0.85.

Which bet is​ better, a $1.14 bet in the Pick 3 game or a $ 1.14 bet in the Pick 4​ game? Explain.

A. The Pick 3 game is a better bet because it has a larger expected value.

B. Neither bet is better because both games have the same expected value.

C.the Pick 4 game is a better bet because it has a larger expected value.

Answer 1

Binomial distribution formula that is to be used for 1. and 2. is: P(X=r) =C(n, r) p^r q^n-r

where, n= number of trials or sample size; p =probability of success on single trial; q =probability of failure on single trial =1-p; C(n, r) =nC_r

1.

The probability of no correct answers =P(X=0) =8C₀ (0.5)⁰ (0.5)⁸ =1(1)(0.0039) =0.0039

2.

A.

Given: n= 11; probability of success, p =0.78; probability of failure, q =1 - 0.78 =0.22

The probability that at least 10 of them need correction for their eyesight =P(X$\geq$10) =P(X=10)+P(X=11)

=11C₁₀ (0.78)^{​​​​​10}(0.22)¹ +11C₁₁ (0.78)¹¹(0.22)⁰

=0.2017+0.0650=0.2667

B.

P(X=10) =0.2017 > 0.05.

So, 10 is a significantly high number of adults requiring eyesight​ correction.

So, option c. is correct. "Yes​, because the probability of this occurring is not small".

3.

a.

n= 10; r =3 with repetition.

Number of different selections that are possible

=(n+r-1)C_r =12C₃ =220

b.

The probability of​ winning =1/220 =0.0045

c.

Net profit on winning =287.33 - 1.14 =$286.19

d.

The expected value, E(X) =287.33(1/220) - 1.14(219/120)

=$0. 17

e.

Expected value =E(Y) = -$0.85

Thus, option A. is correct. "The Pick 3 game is a better bet because it has a larger expected value". 0.17> - 0.85.