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

The probability is:

0.9818

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</p><ul><li></li></ul><p>[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`