A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is:

Question

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is: (A) (17/35) (B) (13/35) (C) (11/35) (D) (10/35)

Tardigrade Admin · Accepted Answer

p = p (correct answer), q = p (wrong answer)

\Rightarrow π = \frac{1}{3}, q = \frac{2}{3}, n = 5

By using Binomial distribution
Required probability

=^{5} C_{4} (\frac{1}{3})^{4} \cdot \frac{2}{3} +^{5} C_{5} (\frac{1}{3})^{5}

= 5 \cdot \frac{2}{3 ^{5}} + \frac{1}{3 ^{5}} = \frac{11}{3 ^{5}}

Q. A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is:

$\frac{17}{3 ^{5}}$

$\frac{13}{3 ^{5}}$

$\frac{11}{3 ^{5}}$

$\frac{10}{3 ^{5}}$

p = p (correct answer), q = p (wrong answer)
$\Rightarrow π = \frac{1}{3}, q = \frac{2}{3}, n = 5$
By using Binomial distribution
Required probability
$=^{5} C_{4} (\frac{1}{3})^{4} \cdot \frac{2}{3} +^{5} C_{5} (\frac{1}{3})^{5}$
$= 5 \cdot \frac{2}{3 ^{5}} + \frac{1}{3 ^{5}} = \frac{11}{3 ^{5}}$

Q. A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is:

3517​

3513​

3511​

3510​

p = p (correct answer), q = p (wrong answer) ⇒π=31​,q=32​,n=5 By using Binomial distribution Required probability =5C4​(31​)4⋅32​+5C5​(31​)5 =5⋅352​+351​=3511​

$\frac{17}{3 ^{5}}$

$\frac{13}{3 ^{5}}$

$\frac{11}{3 ^{5}}$

$\frac{10}{3 ^{5}}$

p = p (correct answer), q = p (wrong answer)
$\Rightarrow π = \frac{1}{3}, q = \frac{2}{3}, n = 5$
By using Binomial distribution
Required probability
$=^{5} C_{4} (\frac{1}{3})^{4} \cdot \frac{2}{3} +^{5} C_{5} (\frac{1}{3})^{5}$
$= 5 \cdot \frac{2}{3 ^{5}} + \frac{1}{3 ^{5}} = \frac{11}{3 ^{5}}$