Math / Data & Statistics

QuestionJaymes is taking a 19-question science test. a. If Jaymes guesses on each question, and the test contains 19 true-false questions, find the probability that he answers 9 questions correctly. $\square$ State answer as a decimal rounded to six decimal places. b. If Jaymes guesses on each question, and the test contains 19 multiple-choice questions, find the probability that he answers 9 questions correctly when each question has 4 answer options. $\square$ State answer as a decimal rounded to six decimal places.

Studdy Solution

STEP 1

What is this asking? What's the chance Jaymes gets exactly 9 questions right on a 19-question test if he's guessing, first for true/false, then for multiple choice with 4 options? Watch out! Don't mix up the formulas for true/false and multiple choice!
Also, remember we want exactly 9 correct answers, not at least 9.

STEP 2

1. True/False Probability
2. Multiple Choice Probability

STEP 3

Alright, let's dive into the true/false part!
We've got 19 questions, and Jaymes is taking a wild guess on each one.
The probability of getting a single true/false question right by guessing is $\frac{1}{2}$ , and the probability of getting it wrong is also $\frac{1}{2}$ .

STEP 4

We want to find the probability of getting exactly 9 questions right.
This is a binomial probability problem!
The formula for binomial probability is: $P(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$ Where $n$ is the number of trials (19 questions), $k$ is the number of successes (9 correct answers), and $p$ is the probability of success on a single trial ( $\frac{1}{2}$ for guessing correctly).

STEP 5

Let's plug in our values: $n = 19$ , $k = 9$ , and $p = \frac{1}{2}$ .
So, we have: $P(X=9) = \binom{19}{9} \cdot \left(\frac{1}{2}\right)^9 \cdot \left(1-\frac{1}{2}\right)^{19-9}$

STEP 6

Now, let's calculate!
First, the binomial coefficient: $\binom{19}{9} = \frac{19!}{9!(19-9)!} = \frac{19!}{9!10!} = 92378$ Then, simplify the rest: $\left(\frac{1}{2}\right)^9 = \frac{1}{512}$ $\left(1-\frac{1}{2}\right)^{19-9} = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$

STEP 7

Putting it all together: $P(X=9) = 92378 \cdot \frac{1}{512} \cdot \frac{1}{1024} = \frac{92378}{524288} \approx 0.176261$

STEP 8

Now for the multiple-choice fun!
We still have 19 questions, but this time, each question has 4 options.
The probability of guessing correctly is $\frac{1}{4}$ , and the probability of guessing incorrectly is $\frac{3}{4}$ .

STEP 9

We're using the same binomial probability formula, but with different values: $n = 19$ , $k = 9$ , and $p = \frac{1}{4}$ . $P(X=9) = \binom{19}{9} \cdot \left(\frac{1}{4}\right)^9 \cdot \left(1-\frac{1}{4}\right)^{19-9}$

STEP 10

We already know $\binom{19}{9} = 92378$ .
Let's calculate the other parts: $\left(\frac{1}{4}\right)^9 = \frac{1}{262144}$ $\left(1-\frac{1}{4}\right)^{19-9} = \left(\frac{3}{4}\right)^{10} \approx 0.0563135$

STEP 11

Combine everything: $P(X=9) = 92378 \cdot \frac{1}{262144} \cdot \left(\frac{3}{4}\right)^{10} \approx 92378 \cdot \frac{1}{262144} \cdot 0.0563135 \approx 0.198322$

STEP 12

a. The probability of Jaymes getting exactly 9 true/false questions correct is approximately 0.176261. b. The probability of Jaymes getting exactly 9 multiple-choice questions correct is approximately 0.198322.

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Studdy Solution

STEP 1

STEP 2

1. True/False Probability
2. Multiple Choice Probability

STEP 3

Alright, let's dive into the true/false part!
We've got 19 questions, and Jaymes is taking a wild guess on each one.
The probability of getting a single true/false question right by guessing is $\frac{1}{2}$ , and the probability of getting it wrong is also $\frac{1}{2}$ .

STEP 4

STEP 5

Let's plug in our values: $n = 19$ , $k = 9$ , and $p = \frac{1}{2}$ .
So, we have: $P(X=9) = \binom{19}{9} \cdot \left(\frac{1}{2}\right)^9 \cdot \left(1-\frac{1}{2}\right)^{19-9}$

STEP 6

Now, let's calculate!
First, the binomial coefficient: $\binom{19}{9} = \frac{19!}{9!(19-9)!} = \frac{19!}{9!10!} = 92378$ Then, simplify the rest: $\left(\frac{1}{2}\right)^9 = \frac{1}{512}$ $\left(1-\frac{1}{2}\right)^{19-9} = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024}$

STEP 7

Putting it all together: $P(X=9) = 92378 \cdot \frac{1}{512} \cdot \frac{1}{1024} = \frac{92378}{524288} \approx 0.176261$

STEP 8

Now for the multiple-choice fun!
We still have 19 questions, but this time, each question has 4 options.
The probability of guessing correctly is $\frac{1}{4}$ , and the probability of guessing incorrectly is $\frac{3}{4}$ .

STEP 9

We're using the same binomial probability formula, but with different values: $n = 19$ , $k = 9$ , and $p = \frac{1}{4}$ . $P(X=9) = \binom{19}{9} \cdot \left(\frac{1}{4}\right)^9 \cdot \left(1-\frac{1}{4}\right)^{19-9}$

STEP 10

We already know $\binom{19}{9} = 92378$ .
Let's calculate the other parts: $\left(\frac{1}{4}\right)^9 = \frac{1}{262144}$ $\left(1-\frac{1}{4}\right)^{19-9} = \left(\frac{3}{4}\right)^{10} \approx 0.0563135$

STEP 11

Combine everything: $P(X=9) = 92378 \cdot \frac{1}{262144} \cdot \left(\frac{3}{4}\right)^{10} \approx 92378 \cdot \frac{1}{262144} \cdot 0.0563135 \approx 0.198322$

STEP 12

a. The probability of Jaymes getting exactly 9 true/false questions correct is approximately 0.176261. b. The probability of Jaymes getting exactly 9 multiple-choice questions correct is approximately 0.198322.

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Studdy Solution

STEP 1

STEP 2

1. True/False Probability2. Multiple Choice Probability

STEP 3

Alright, let's dive into the true/false part!We've got 19 questions, and Jaymes is taking a wild guess on each one.The probability of getting a single true/false question right by guessing is 12 \frac{1}{2} 21​, and the probability of getting it wrong is also 12 \frac{1}{2} 21​.

STEP 4

STEP 5

Let's plug in our values: n=19n = 19n=19, k=9k = 9k=9, and p=12p = \frac{1}{2}p=21​.So, we have: P(X=9)=(199)⋅(12)9⋅(1−12)19−9 P(X=9) = \binom{19}{9} \cdot \left(\frac{1}{2}\right)^9 \cdot \left(1-\frac{1}{2}\right)^{19-9} P(X=9)=(919​)⋅(21​)9⋅(1−21​)19−9

STEP 6

STEP 7

Putting it all together: P(X=9)=92378⋅1512⋅11024=92378524288≈0.176261 P(X=9) = 92378 \cdot \frac{1}{512} \cdot \frac{1}{1024} = \frac{92378}{524288} \approx 0.176261 P(X=9)=92378⋅5121​⋅10241​=52428892378​≈0.176261

STEP 8

Now for the multiple-choice fun!We still have 19 questions, but this time, each question has 4 options.The probability of guessing correctly is 14 \frac{1}{4} 41​, and the probability of guessing incorrectly is 34 \frac{3}{4} 43​.

STEP 9

STEP 10

STEP 11

STEP 12

a. The probability of Jaymes getting exactly 9 true/false questions correct is approximately **0.176261**. b. The probability of Jaymes getting exactly 9 multiple-choice questions correct is approximately **0.198322**.

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1. True/False Probability
2. Multiple Choice Probability

Alright, let's dive into the true/false part!
We've got 19 questions, and Jaymes is taking a wild guess on each one.
The probability of getting a single true/false question right by guessing is $\frac{1}{2}$ , and the probability of getting it wrong is also $\frac{1}{2}$ .

Let's plug in our values: $n = 19$ , $k = 9$ , and $p = \frac{1}{2}$ .
So, we have: $P(X=9) = \binom{19}{9} \cdot \left(\frac{1}{2}\right)^9 \cdot \left(1-\frac{1}{2}\right)^{19-9}$

Putting it all together: $P(X=9) = 92378 \cdot \frac{1}{512} \cdot \frac{1}{1024} = \frac{92378}{524288} \approx 0.176261$

Now for the multiple-choice fun!
We still have 19 questions, but this time, each question has 4 options.
The probability of guessing correctly is $\frac{1}{4}$ , and the probability of guessing incorrectly is $\frac{3}{4}$ .

a. The probability of Jaymes getting exactly 9 true/false questions correct is approximately 0.176261. b. The probability of Jaymes getting exactly 9 multiple-choice questions correct is approximately 0.198322.