A question paper contains 90 multiple-choice questions. Each correct answer carries 1 mark. There are 4 alternative answers (A, B, C or D), of which only one is correct. Mr X answers these questions randomly (i.e. without preparation). What is the probability that X gets at least 10 marks?
Question
A question paper contains 90 multiple-choice questions. Each correct answer carries 1 mark. There are 4 alternative answers (A, B, C or D), of which only one is correct. Mr X answers these questions randomly (i.e. without preparation). What is the probability that X gets at least 10 marks?
Solution
To solve this problem, we need to use the concept of binomial probability. The binomial probability formula is:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success on a single trial
- n is the number of trials
- k is the number of successes
In this case:
- n = 90 (the number of questions)
- p = 1/4 (the probability of getting a question right by guessing)
- k = 10 (the number of correct answers we're interested in)
However, we're interested in the probability of getting at least 10 correct answers, so we need to calculate the probability of getting 10, 11, 12, ..., 90 correct answers and sum these probabilities.
This calculation can be quite complex, so it's easier to calculate the probability of getting 0 to 9 correct answers and subtract this from 1.
Let's calculate the probability of getting k correct answers for k = 0 to 9 and sum these probabilities:
P = P(X=0) + P(X=1) + P(X=2) + ... + P(X=9)
Then, the probability of getting at least 10 correct answers is:
P(at least 10) = 1 - P
This calculation involves a lot of computation, so it's best to use a statistical software or a calculator with statistical functions to get the exact result.
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