Answer to Question #331049 in Statistics and Probability for jaz

Question

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=.65 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

n = 7

P(success) = 0.65

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:

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

$P(X=0)=C(7,0)\cdot 0.65^0\cdot(1-0.65)^{7-0}=\frac{7!}{0!(7-0)!}\cdot0.65^0\cdot0.35^7=0.0006$

$P(X=1)=C(7,1)\cdot 0.65^1\cdot(1-0.65)^{7-1}=\frac{7!}{1!(7-1)!}\cdot0.65^1\cdot0.35^6=0.0083$

$P(X=2)=C(7,2)\cdot 0.65^2\cdot(1-0.65)^{7-2}=\frac{7!}{2!(7-2)!}\cdot0.65^2\cdot0.35^5=0.0466$

$P(X=3)=C(7,3)\cdot 0.65^3\cdot(1-0.65)^{7-3}=\frac{7!}{3!(7-3)!}\cdot0.65^3\cdot0.35^4=0.1442$

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

$P(X<4)=0.0006+0.0083+0.0466+0.1442=0.1997$