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Zepler [3.9K]
3 years ago
5

Richard has just been given an l0-question multiple-choice quiz in his history class. Each question has five answers, of which o

nly one is correct, Since Richard has not attended a class recently, he doesn't know any of the answers, Assuming that Richard guesses on all 10 questions. Find the indicated probabilities.
A) What is the probability that he will answer all questions correctly?
B) What is the probability that he will answer all questions incorrectly?
C) What is the probability that he will answer at least one of the questions correctly?
Then use the fact that P(r1) = 1 P(r = 0).
D) What is the probability that Richard will answer at least half the questions correctly?
Mathematics
1 answer:
myrzilka [38]3 years ago
8 0

Answer:

a) 0.0000001024 probability that he will answer all questions correctly.

b) 0.1074 = 10.74% probability that he will answer all questions incorrectly

c) 0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

d) 0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

Step-by-step explanation:

For each question, there are only two possible outcomes. Either he answers it correctly, or he does not. The probability of answering a question correctly is independent of any other question. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Each question has five answers, of which only one is correct

This means that the probability of correctly answering a question guessing is p = \frac{1}{5} = 0.2

10 questions.

This means that n = 10

A) What is the probability that he will answer all questions correctly?

This is P(X = 10)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} = 0.0000001024

0.0000001024 probability that he will answer all questions correctly.

B) What is the probability that he will answer all questions incorrectly?

None correctly, so P(X = 0)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074

0.1074 = 10.74% probability that he will answer all questions incorrectly

C) What is the probability that he will answer at least one of the questions correctly?

This is

P(X \geq 1) = 1 - P(X = 0)

Since P(X = 0) = 0.1074, from item b.

P(X \geq 1) = 1 - 0.1074 = 0.8926

0.8926 = 89.26% probability that he will answer at least one of the questions correctly.

D) What is the probability that Richard will answer at least half the questions correctly?

This is

P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 5) = C_{10,5}.(0.2)^{5}.(0.8)^{5} = 0.0264

P(X = 6) = C_{10,6}.(0.2)^{6}.(0.8)^{4} = 0.0055

P(X = 7) = C_{10,7}.(0.2)^{7}.(0.8)^{3} = 0.0008

P(X = 8) = C_{10,8}.(0.2)^{8}.(0.8)^{2} = 0.0001

P(X = 9) = C_{10,9}.(0.2)^{9}.(0.8)^{1} \approx 0

P(X = 10) = C_{10,10}.(0.2)^{10}.(0.8)^{0} \approx 0

So

P(X \geq 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0264 + 0.0055 + 0.0008 + 0.0001 + 0 + 0 = 0.0328

0.0328 = 3.28% probability that Richard will answer at least half the questions correctly

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Step-by-step explanation:

(a) You bet $3 on a single round which means that if you win the game, your amount will double ($6), your profit will be $3. Whereas, if you lose the round, your profit will be -$3. You can only bet on red or black and both have 18 slots each.

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S.D = √Var(x)

We need to first calculate E(x²).

E(x²) = ∑x².P(x)

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Var(x) = E(x²) - E(x)²

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(b) Now, the betting price is $1 and 3 rounds are played. We will compute the expectation for one round and then add it thrice to find the expectation for three rounds. Similarly, for the standard deviations, we will add the individual variances and then consider the square root of it.

E(x) = ∑ x.P(x)

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E(x) = -0.027

Standard deviation can be calculated by the following formula:

Var(x) = E(x²) - E(x)²

S.D = √Var(x)

We need to first calculate E(x²).

E(x²) = ∑x².P(x)

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       = (1)(18/37) + (1)(19/37)

E(x²) = 1

Var(x) = E(x²) - E(x)²

         = 1 - (-0.027)²

Var(x) = 0.9992

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E(x₁ + x₂ + x₃) = E(x₁) + E(x₂) + E(x₃)

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E(x₁ + x₂ + x₃) = -0.081

Similarly, the variance for one round is 0.9992.

Var (x₁ + x₂ + x₃) = Var(x₁) + Var(x₂) + Var(x₃)

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Var (x₁ + x₂ + x₃) = 2.9976

S.D = √2.9976

S.D = 1.73              

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