Question

Stan guessed on all 10 questions of a multiple-choice quiz. Each question has 4 answer choices. What is the probability that he got at least 2 questions correct? Round the answer to the nearest thousandth.

Answer 1

Explanation:

C on edge Test

Answer 2

The probability that he got at least 2 questions correct is 0.756

Lets assume, the event of getting a question correct is P

As, each question has 4 answer choices and only one choice is correct,

So, P =
`(1)/(4)`

When
the probability of getting 1 success in 1 trial is p, then the probability of getting exactly x successes out of n trials is given by the
formula :

(ⁿCₓ )(P)ˣ (1-P)ⁿ⁻ˣ

Here the total number of questions is 10. So, n= 10

Stan needs to get at least 2 questions correct, that means he can get 2, 3, 4, 5, 6, 7, 8, 9 or 10 questions correct.

Now according to the formula,

P(x=2) = ¹⁰C₂ (
`(1)/(4)`)² (
`1- (1)/(4)`)¹⁰⁻²

= ¹⁰C₂
`((1)/(16))((3)/(4))^8`

= 0.281567573

P(x=3) = ¹⁰C₃ (
`(1)/(4)`)³
`((3)/(4))^7`

= (120) (
`(1)/(4)`)³
`((3)/(4))^7`

= 0.250282287

P(x=4) = ¹⁰C₄ (
`(1)/(4)`)⁴
`((3)/(4))^6`

= (210)(
`(1)/(4)`)⁴
`((3)/(4))^6`

= 0.145998001

P(x=5) = ¹⁰C₅ (
`(1)/(4)`)⁵
`((3)/(4))^5`

= (252) (
`(1)/(4)`)⁵
`((3)/(4))^5`

= 0.058399200

P(x=6) = ¹⁰C₆ (
`(1)/(4)`)⁶
`((3)/(4))^4`

= (210) (
`(1)/(4)`)⁶
`((3)/(4))^4`

= 0.016222000

P(x=7) = ¹⁰C₇ (
`(1)/(4)`)⁷
`((3)/(4))^3`

= (120) (
`(1)/(4)`)⁷
`((3)/(4))^3`

= 0.003089904

P(x=8) = ¹⁰C₈ (
`(1)/(4)`)⁸
`((3)/(4))^2`

= (45) (
`(1)/(4)`)⁸
`((3)/(4))^2`

= 0.000386238

P(x=9) = ¹⁰C₉ (
`(1)/(4)`)⁹
`((3)/(4))^1`

= (10) (
`(1)/(4)`)⁹
`((3)/(4))^1`

= 0.000028610

P(x=10) = ¹⁰C₁₀ (
`(1)/(4)`)¹⁰

= 0.000000953

Now we will just add all the probabilities and get:

0.281567573 + 0.250282287+ 0.145998001+ 0.058399200+ 0.016222000+0.003089904+0.000386238+ 0.000028610 +0.000000953

= 0.755974766

= 0.756 (rounding to the nearest thousandth)

So, the probability that he got at least 2 questions correct is 0.756