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 .

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.

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )

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step-by-step explanation: