Answer to Question #147155 in Statistics and Probability for Vladimyr Lubin

Question

(5.2.17) 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=0.35. Find the indicated probability for the number of correct answers.
Find the probability that the number x of correct answers is fewer than 4.
P(X<4)=? (Round to four decimal)

Answer 1

n = 7

P(success) = 0.35

The number of successes among a fixed number of independent trials follows binomial distribution. Lets evaluate the definition of binomial probability at m = 0, 1, 2, 3:

$$P(X=m)=C(n,m)\cdot p^m\cdot(1-p)^{n-m}$$

$P(X=0)=C(7,0)\cdot 0.35^0\cdot(1-0.35)^{7-0}=\frac{7!}{0!(7-0)!}\cdot0.35^0\cdot0.65^7=0.049$

$P(X=1)=C(7,1)\cdot 0.35^1\cdot(1-0.35)^{7-1}=\frac{7!}{1!(7-1)!}\cdot0.35^1\cdot0.65^6=0.1848$

$P(X=2)=C(7,2)\cdot 0.35^2\cdot(1-0.35)^{7-2}=\frac{7!}{2!(7-2)!}\cdot0.35^2\cdot0.65^5=0.2985$

$P(X=3)=C(7,3)\cdot 0.35^3\cdot(1-0.35)^{7-3}=\frac{7!}{3!(7-3)!}\cdot0.35^3\cdot0.65^4=0.2679$

$P(X<4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$

$P(X<4)=0.049+0.1848+0.2985+0.2679=0.8002$