Question

Assume that random guesses are made for 15 multiple chioce questions on an SAT test and that there are 5 choices on each question with the probability of success 0.20. Find the probability that the number of correct answers is at most 6 (This problem meets all the requirements of a binomial situation.)

Answer 1

Answer:

The probability is:

                          0.9818

Step-by-step explanation:

We need to use the binomial theorem in order to find the probability.

We know that when there are r successes out of n total experiments such that p denote the probability of success then the probability of r successes is given by the formula:

$P(X=r)=n_C_r\cdot p^r\cdot (1-p)^{n-r}$

Here we khave:

n=15

p=0.20

Hence,

1-p=1-0.20=0.80

Also, we are asked to find the probability that the number of correct answers is at most 6 i.e.

$P(X\leq 6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)$

$P(X=0)=15_C_0\cdot (0.20)^{0}\cdot (0.80)^{15-0}\\\\i.e.\\\\P(X=1)=1\cdot 1\cdot (0.80)^{15}\\\\i.e.\\\\P(X=1)=0.0351

[tex]P(X=1)=15_C_1\cdot (0.20)^{1}\cdot (0.80)^{15-1}\\\\i.e.\\\\P(X=1)=15\cdot (0.20)\cdot (0.80)^{14}\\\\i.e.\\\\P(X=1)=0.1319$

$P(X=2)=15_C_2\cdot (0.20)^{2}\cdot (0.80)^{15-2}\\\\i.e.\\\\P(X=2)=15_C_2\cdot (0.20)^{2}\cdot (0.80)^{13}\\\\i.e.\\\\P(X=2)=0.2309$

$P(X=3)=15_C_3\cdot (0.20)^{3}\cdot (0.80)^{15-3}\\\\i.e.\\\\P(X=3)=15_C_3\cdot (0.20)^{3}\cdot (0.80)^{12}\\\\i.e.\\\\P(X=3)=0.2501$

$P(X=4)=15_C_4\cdot (0.20)^{4}\cdot (0.80)^{15-4}\\\\i.e.\\\\P(X=4)=15_C_4\cdot (0.20)^{4}\cdot (0.80)^{11}\\\\i.e.\\\\P(X=4)=0.1876$

$P(X=5)=15_C_5\cdot (0.20)^{5}\cdot (0.80)^{15-5}\\\\i.e.\\\\P(X=5)=15_C_5\cdot (0.20)^{5}\cdot (0.80)^{10}\\\\i.e.\\\\P(X=5)=0.1032$

$P(X=6)=15_C_6\cdot (0.20)^{6}\cdot (0.80)^{15-6}\\\\i.e.\\\\P(X=6)=15_C_6\cdot (0.20)^{6}\cdot (0.80)^{9}\\\\i.e.\\\\P(X=6)=0.043$

Hence,

$P(X\leq 6)=0.0351+0.1319+0.2309+0.2501+0.1876+0.1032+0.043\\\\i.e.\\\\P(X\leq 6)=0.9818$