In an examination a candidate is given the four answers to four questions but is not told which answer applies to which question. He is asked to write down each of the four answers next to its appropriate question.
b). show that there are 6 ways of getting just two answers in the correct places.
c). If a candidate guesses at random where the four answers are to go and X is the number of correct guesses he makes, draw up the probability distribution for X in tabular form.
d). calculate the expectations of X.
b). To show that there are 6 ways of getting just two answers in the correct places, we can use the formula for permutations.
There are 4 questions and we want exactly 2 answers to be in the correct place. So, we need to choose 2 out of the 4 questions to have their answers in the correct place.
The number of ways to choose 2 questions out of 4 is given by the combination formula:
C(4, 2) = 4! / (2! * (4-2)!) = 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6
Therefore, there are 6 ways of getting just two answers in the correct places.
c). If a candidate guesses at random where the four answers are to go and X is the number of correct guesses he makes, the probability distribution for X can be represented in tabular form as follows:
X | Probability
0 | P(0)
1 | P(1)
2 | P(2)
3 | P(3)
4 | P(4)
The probability of getting 0 correct guesses (P(0)) can be calculated as:
P(0) = (3/4) * (2/3) * (1/2) * (1/1) = 1/4
The probability of getting 1 correct guess (P(1)) can be calculated as:
P(1) = (1/4) * (2/3) * (1/2) * (1/1) = 1/12
The probability of getting 2 correct guesses (P(2)) can be calculated as:
P(2) = (1/4) * (1/3) * (1/2) * (1/1) = 1/24
The probability of getting 3 correct guesses (P(3)) can be calculated as:
P(3) = (1/4) * (1/3) * (1/2) * (1/1) = 1/24
The probability of getting 4 correct guesses (P(4)) can be calculated as:
P(4) = (1/4) * (1/3) * (1/2) * (1/1) = 1/24
d). The expectation of X (E(X)) can be calculated by multiplying each value of X with its probability and summing up the results.
E(X) = 0 * P(0) + 1 * P(1) + 2 * P(2) + 3 * P(3) + 4 * P(4)
E(X) = 0 * (1/4) + 1 * (1/12) + 2 * (1/24) + 3 * (1/24) + 4 * (1/24)
E(X) = 0 + 1/12 + 1/12 + 1/8 + 1/6
E(X) = 1/12 + 1/12 + 1/8 + 1/6
E(X) = 6/72 + 6/72 + 9/72 + 12/72
E(X) = 33/72
Therefore, the expectation of X is 33/72.
To solve parts b), c), and d), we can use the concept of permutations and the probability theory.
a) Total Number of Ways:
To find the total number of ways to arrange the answers, we use the concept of permutations. Since there are 4 questions and 4 answers, the total number of ways is given by 4! (or 4 factorial), which is equal to 4 x 3 x 2 x 1 = 24.
b) Number of Ways to Get Only 2 Correct:
To determine the number of ways to get exactly 2 answers in the correct places, we can use combinations. We choose 2 positions (questions) out of 4 where the candidate gets the answers correct. The remaining 2 positions will have the answers placed incorrectly. The number of ways to choose 2 positions from 4 is given by 4C2 (or 4 choose 2), which is equal to 4! / (2! x (4 - 2)!) = 6.
c) Probability Distribution for X:
If the candidate guesses randomly, the probability of getting a particular question correct is 1/4, as there are 4 possible answers to choose from for each question. We can calculate the probabilities of different values of X (number of correct guesses) using the binomial distribution formula.
The probability distribution for X can be represented in tabular form as follows:
```
X (number of correct guesses) Probability (P(X))
0 (3/4)^4
1 4C1 * (1/4)^1 * (3/4)^3
2 4C2 * (1/4)^2 * (3/4)^2
3 4C3 * (1/4)^3 * (3/4)^1
4 (1/4)^4
```
d) Expectation of X:
To calculate the expectation of X, we need to multiply each value of X by its corresponding probability and sum them up.
```
Expectation of X = (0 * P(X=0)) + (1 * P(X=1)) + (2 * P(X=2)) + (3 * P(X=3)) + (4 * P(X=4))
```
You can substitute the probabilities calculated in part c) into the expectation formula to find the numerical value.