Question

assume that random guesses are made for 8 multiple choice questions on an SAT test, so that there are n=9 trials, each with probability of success (correct) given by p=0.6. find the indicated probability for the number of correct answers. find the probability that the number x of correct answers is fewer than 4.

Answer 1

Answer:

$P(x<4)=0.0994$

Step-by-step explanation:

If we call x the number of correct questions obtained in the 9 attempts, then:

x is a discrete random variable that can be modeled by a binomial probability distribution p, with n = 9 trials.

So, the p of x successes has the following formula.

$P(x) =\frac{n!}{x!(n-x)!}*p^x(1-p)^{n-x}$

Where:

n = 9

p = 0.6

We are looking for P(x<4)

By definition:

$P(x<4) = P(x\leq3) = P(0) + P(1) + P(2) +P(3)$

Then:

$P(x\leq3)=\sum_{x=0}^{3} \frac{9!}{x!(9-x)!}*(0.6)^x(1-0.6)^{9-x}$

$P(x\leq3)=0.0994$

Answer 2

Probabilities are used to determine the chances of an event

The probability that the number of correct answers is fewer than 4 is 0.0994

The given parameters are:

$n = 9$

$p =0.6$

The probability is a binomial probability, and it is calculated using:

$P(x) = ^nC_x p^x (1 - p)^{n -x}$

Fewer than 4 means: x = 0, 1, 2 and 3

So, we have:

$P(x<4) = ^9C_0 \times 0.6^0 \times (1 - 0.6)^{9 -0} +^9C_1 \times 0.6^1 \times(1 - 0.6)^{9 -1} +^9C_2 \times 0.6^2\times (1 - 0.6)^{9 -2} +^9C_3 \times 0.6^3\times (1 - 0.6)^{9 -3}$

This gives

$P(x<4) = 1 \times 0.6^0 \times (1 - 0.6)^9 + 9 \times 0.6^1 \times(1 - 0.6)^8 +36 \times 0.6^2 \times (1 - 0.6)^7 +84 \times 0.6^3 \times (1 - 0.6)^6$

Using a calculator,

$P(x<4) = 0.099352576$

Approximate

$P(x<4) = 0.0994$

Hence, the probability that the number of correct answers is fewer than 4 is 0.0994

Read more about binomial probabilities at:

https://brainly.com/question/23498586