Assume that random guesses are made for seven multiple choice questions on an SAT​ test, so that there are n=7 ​trials, each with probability of success​ (correct) given by p=0.55. Find the indicated probability for the number of correct answers. Find the probability that the number x of correct answers is fewer than 4.

Let the random variable $X$ represent the random guesses then,  $X\sim Binomial(n=7,p=0.55)$.  

The probability that the number x of correct answers is fewer than 4 is given as,

$p(X\lt4)=p(X=0)+p(X=1)+p(X=2)+p(X=3)$

Now,

$p(X=0)=\binom{7}{0}0.55^00.45^7=0.45^7=0.0037\\ p(X=1)=\binom{7}{1}0.55^10.45^6=7\times 0.55\times0.0083=0.0319695\\ p(X=2)=\binom{7}{2}0.55^20.45^5=21\times0.3025\times 0.0185=0.11722148\\ p(X=3)=\binom{7}{3}0.55^30.45^4=35\times0.166375\times0.0410=0.23878452$

Therefore,

$p(X\lt 4)=0.0037+0.0319695+0.11722148+0.23878452= 0.3917122$

The probability that the number $x$ of correct answers is fewer than 4 is 0.3917122