contestada

Assume that random guesses are made for seven multiple choice questions on an SAT​ test, so that there are nequals 7 ​trials, each with probability of success​ (correct) given by pequals 0.25 . Find the indicated probability for the number of correct answers. Find the probability that the number x of correct answers is fewer than 4 .

Respuesta :

Answer:

0.929.

Step-by-step explanation:

The event x fewer than 4 is 0, 1 2 or 3 correct answers.

Probability of the first 3 being correct and the last 4 incorrect

= (0.25)^3 * (0.75)^4 = 0.00494385

There are  7C3 = 7*6*5/ 3*2*1 = 35 ways in which this can happen so the required probability for 3 correct answers is 35 * 0.00494385 = 0.17303.

In a similar fashion we can find the probability of  0, 1 and 2 successes:

P(0) = (0.75)^7 =  0.13348

P(1) =  7 * 0.25*(0.75)^6 = 0.31146

P(2) = 21* (0.25)^2(0.75)^5 = 0.31146

So probability of 1 2 or 3 successes = sum of the above 4 probabilities

=  929.

fichoh

The probability that the number of correct answers in the SAT test is fewer than 4 is 0.9295

Given the Parameters :

  • Probability of success, p = 0.25
  • Number of trials, n = 7
  • X < 4
  • q = 1 - p = 1 - 0.25 = 0.75

Using the concept of binomial probability defined as :

P(x = x) = nCx * p^x * q^(n-x)

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

Using a binomial probability calculator :

  • P(x = 0) = 0.1335
  • P(x = 1) = 0.3115
  • P(x = 2) = 0.3115
  • P(x = 3) = 0.1730

P(x < 4) = 0.1335 + 0.3115 + 0.3115 +0.1730 = 0.9295

Therefore, the probability that number of correct answers is fewer than 4 is 0.9295

Learn more :https://brainly.com/question/12474772

Otras preguntas