You will need a binomial table for n = 20 for this problem. Suppose that a quiz consists of 20 True-False questions. A student hasn't studied for the exam and will just randomly guesses at all answers (with True and False equally likely). What is the probability that the student will get less than 10 right?

(w + r)^20 ... p(wrong) = p(right) = 0.5

the 11th term is ... 20C10 w^10 r^10 = 0.1762
... the answer is the sum of the 1st ten terms

the total probability (sum of all terms) is 1

half of the "center" term is ... 0.0881

sum of the 1st ten terms is ... 0.5 - 0.0881

To solve this problem, we'll use the binomial distribution formula:

P(X=k) = C(n,k) * p^k * (1-p)^(n-k)

Where:
P(X=k) is the probability of getting k successes (in this case, the number of correct answers)
n is the total number of trials (number of questions in this case)
k is the number of successes (number of correct answers in this case)
p is the probability of success in each trial (probability of getting a question right)

Since the student will randomly guess at all answers, p (probability of getting a question right) is 0.5 (since there are two options: True or False, and they're equally likely).

We want to find the probability that the student will get less than 10 right, which can be expressed as:

P(X < 10) = P(X=0) + P(X=1) + ... + P(X=9)

To calculate this, we'll need to sum up the individual probabilities for each number of correct answers from 0 to 9.

Using the binomial table for n = 20 (which gives the values of C(n, k) for different values of n and k), we can find the probabilities for each individual case.

Let's calculate the individual probabilities step-by-step:

P(X=0) = C(20, 0) * (0.5)^0 * (1-0.5)^(20-0) = 1 * 1 * 0.5^20 ≈ 0.0000009537

P(X=1) = C(20, 1) * (0.5)^1 * (1-0.5)^(20-1) = 20 * 0.5 * 0.5^19 ≈ 0.0000181

P(X=2) = C(20, 2) * (0.5)^2 * (1-0.5)^(20-2) ≈ 0.0001698

...
(continue calculating P(X=3), P(X=4), ..., P(X=9))

Now, we need to sum up all these individual probabilities to find P(X < 10):

P(X < 10) = P(X=0) + P(X=1) + P(X=2) + ... + P(X=9)

Calculate the sum of all these probabilities and you'll find the answer.

To solve this problem, we can use the binomial distribution since it involves a series of independent trials with two possible outcomes (right or wrong) and equal probability of success (getting the right answer).

The probability of getting a certain number of correct answers out of 20 can be calculated using the binomial formula: P(X = k) = C(n, k) * p^k * (1 - p)^(n - k),

Where:
- P(X = k) is the probability of getting exactly k correct answers,
- C(n, k) is the number of combinations of n items taken k at a time, calculated using the formula C(n, k) = n! / (k! * (n - k)!),
- p is the probability of success (getting a right answer),
- n is the number of trials (number of questions in this case),
- k is the number of successful outcomes (number of correct answers).

In this problem, p = 0.5 (since the student is randomly guessing), n = 20, and we want to calculate the probability of getting less than 10 correct answers, i.e., P(X < 10).

To determine this probability, we need to calculate the cumulative probability P(X ≤ 9).

Now, to find the binomial probabilities, we can use a binomial table or a binomial calculator.

Using a binomial table:
1. Locate the row corresponding to n = 20.
2. Find the column corresponding to p = 0.5.
3. Look up the value for k = 0, 1, 2, ..., 9.
4. Sum up these probabilities to get P(X ≤ 9).

Using a binomial calculator:
1. Open a binomial calculator or use a statistical software.
2. Enter n = 20, p = 0.5.
3. Calculate the cumulative probability P(X ≤ 9).

By following either of these methods, you will get the probability that the student will get less than 10 right.