# Can Someone Shed Some Light On This ProblemWatch

#1
A multiple choice examination consists of 20 questions each of which have 4
possible answers. A candidate gets 5 marks for a correct answer and has 2 marks
deducted for an incorrect answer. (If the candidate does not answer the question
no marks are awarded or deducted.) The pass mark is 30.
a) What are the maximum and minimum possible examination marks? [2]
b) What is the minimum number of questions a candidate needs to answer
correctly in order to pass the examination? [1]
c) Estimate the number of marks a candidate would get if he or she answered
each question by picking 1 of the 4 multiple choice answers at random. [1]
d) Assume a candidate answers every question, and let x be the number of
i) Write down an expression in terms of x for the number of questions he
ii) Write down an expression in terms of x for the number of marks he
obtains. [3]
iii) If the candidate obtained 44 marks, how many answers did he answer
correctly. [3]
iv) How many must he answer correctly to pass the examination? [3]
e) If, instead, 4 marks are awarded for each correctly answered question
i) what is the minimum number of questions a candidate needs to answer
correctly in order to pass the examination? [2]
ii) If a candidate answers all the questions, how many must a candidate
answer correctly to pass the examination? [4]
0
10 years ago
#2
I get this

a) Gets all right - 20*5 = 100 Gets all wrong - -2*20 = -40
c) On average s/he will get 0.25*5 + 0.75*-2 = -1 per question. Therefore -20 expected.
d) i) 20 - x
ii) 5x - 2(20 - x) = 7x - 40
iii) 7x-40 = 44 Therefore 7x = 84 Therefore x = 12
iv) 7x-40=30 7x=70 x=10
e) i) 8, since 8*4 = 32
ii) Using same methd as before, mark = 4x - 2(20-x) = 6x-40 6x-40=30 6x=70 x=11.something Round up as to pass need 30 or more, x=12
0
#3
0.25*5 + 0.75*-2 = -1

should be 0.25*5 + 0.75*-2 = -0.25

any thoughts
0
10 years ago
#4
Yeah, sorry my mistake, so it would be -5 not -20
0
#5
many thanks you are a gentleman
0
#6
A multiple-choice exam consists of 10 questions, each with 5 possible answers. All questions carry equal weight and 60% is the lowest passing score.If a student will be picking answers at random what's the probability of getting a passing score?
Solution 1

OK, the chance of getting any particular question right is 20%. In order to get a score of 60% or more, the student has to get 6, 7, 8, 9, or 10 right.

The number of possibilities of getting 6 right is 1^6 * 5 * 5 * 5 * 5, or 625/(5^10)or 0.000064, or .0064%, or 64 chances in 10,000, or 4 in 625.

For 7 right, it's 1^7 * 5 * 5 * 5, or 125/9765625, or .0000128.

Add the decimals together for 6, 7, 8, 9 and 10 right, and you will have the probability.

I bet it will come out about .013%. Or not - I guess you will have to do the work. )

Solution 2

if you need 60%, you need atleast 6 questions right.
so, what are the ways to get 10, 9, 8, 7, or 6 questions answered right?

10)
there is a 1/5 chance to answer any one questions right, and you need all of them answered right, so (1/5)^10 chance of getting a all 10 right

9)
(1/5)^9 seems to be the intuitive probability for this, however there are 10 ways to get this result (10 different choices for which answer will be wrong)
so the correct probability here is 10*(1/5)^9

8) there are 10 slots, and you choose 2 (so it's a combination, and not a permutation) 10C2 = 10!/((10-2)!*2!)
and we multiply that number by (1/5)^8

I bet you can figure out 6 and 7

now, multiply: P(10)*P(9)...P(6)
0
10 years ago
#7
(Original post by nazz2222)
now, multiply: P(10)*P(9)...P(6)
You don't want the probability that they will all happen, you want the prob that one of them will. So add.
0
#8
do we have to incorporate combinations like when calculating P(8)

8) there are 10 slots, and you choose 2 (so it's a combination, and not a permutation) 10C2 = 10!/((10-2)!*2!)
and we multiply that number by (1/5)^8
0
10 years ago
#9
You need to calculate the number of ways of getting exactly 6 questions right.
Add the number of ways of getting exactly 7 questions right, exactly 8 questions, ..., all 10 questions right. Then divide by the total number of ways of answering the 10 questions.

To calculate the number of ways, you both need to take into account combinations and that you have to get some questions wrong if you're going to get exactly 6 questions right. (You are currently assuming it doesn't matter whether or not you get the "unneeded" questions right).
0
#10
i am afraid i dont understand what you are saying. could you pleaes write down the steps oh how to calculate probablity of getting 8 questions right.

As for the below question which is the right solution, 1 or 2?

multiple-choice exam consists of 10 questions, each with 5 possible answers. All questions carry equal weight and 60% is the lowest passing score.If a student will be picking answers at random what's the probability of getting a passing score?

many thanks
0
10 years ago
#11
Do a web search for Binomial Distribution.

An example that's reasonably close to what you're asking is at http://www.cardiff.ac.uk/maths/teach..._ex/index.html
0
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