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Nov 2008 p6 q7
2950
A fair die has one face numbered 1, one face numbered 3, two faces numbered 5 and two faces numbered 6.
The die is thrown twice. Let \(X\) be the sum of the two scores. The following table shows the possible values of \(X\).
Second throw
First throw
1
3
5
5
6
6
1
2
4
6
6
7
7
3
4
6
8
8
9
9
5
6
8
10
10
11
11
5
6
8
10
10
11
11
6
7
9
11
11
12
12
6
7
9
11
11
12
12
Draw up a table showing the probability distribution of \(X\).
Calculate \(E(X)\).
Find the probability that \(X\) is greater than \(E(X)\).
Solution
(ii) To find the probability distribution of \(X\), count the occurrences of each sum in the table and divide by the total number of outcomes (36). The probabilities are:
\(x\)
2
4
6
7
8
9
10
11
12
\(P(X=x)\)
\(\frac{1}{36}\)
\(\frac{2}{36}\)
\(\frac{5}{36}\)
\(\frac{4}{36}\)
\(\frac{4}{36}\)
\(\frac{4}{36}\)
\(\frac{4}{36}\)
\(\frac{8}{36}\)
\(\frac{4}{36}\)
(iii) The expected value \(E(X)\) is calculated as: