Question

Please help

Answer 1

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

Answer 2

@Michele_Laino @rational please help

Answer 3

A multiple choice examination has 5 questions.

Each question has three alternative answers of which exactly one is correct.
1 out of 3 is the probability of getting it right

The probability that a student will get 4 or more correct answers just by guessing is

P(4 right) + P(5 right)

Answer 4

there are 3 possible answers in each question
the chance of randomly getting 1 question right is 1/3
the chance of randomly getting 4 questions right is (1/3) * (1/3) * (1/3) * (1/3)

Answer 5

*(2/3) for the wrong answer

Answer 6

there are 5 ways to organize the results

rrrrw
rrrwr
rrwrr
rwrrr
wrrrr

so 5 of them and each one is (1/3)^4 * (2/3)^1

Answer 7

binomial thrm have you covered it yet?

Answer 8

i think its binomial distribution\[P(X \ge4)=nCx\ p^n\ q^{n-x}\]

Answer 9

thats correct

Answer 10

4 correct at 1 in 3
1 wrong at 2 in 3

5 C 4 ways to organise the results

Answer 11

p=1,q=2
n=5

Answer 12

no, p = 1/3 since 1 out of 3 is correct option
q = 1-p , or the probability of a wrong choice

n is 5, x is the random variable of 4

Answer 13

yep
sorry
just forgot the things :)

Answer 14

it happens :)

Answer 15

so am i right @amistre64 ?

Answer 16

right about what? ive already corrected your errors

Answer 17

tell me the values of :

p
q
n
x

Answer 18

hmm.
nothing
p=1/3 q=2/3 n=5 x=4

Answer 19

good

Answer 20

and for P(X=5) is just making x=5

Answer 21

yep

Answer 22

\[P(X=4):=\binom{5}{4}(\frac13)^{4}(\frac23)^{1}\]
\[P(X=5:)=\binom{5}{5}(\frac13)^{5}(\frac23)^{0}\]
\[P(X\ge4):=P(X=4)+P(X=5)\]

Answer 23

yes
thank you so much guys

Answer 24

got a : going astray :)

:= means, "is defined as"

Answer 25

well ill solve it thank u:)