A fair die is thrown two times then the probability that sum of the numbers on them is 5 is

WAEC 2016

A fair die is thrown two times. What is the probability that the sum of the scores is at least 10?

\(\begin{array}{c|c} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\) From the table above, event space, n(E) = 6 sample space, n(S) = 36 Hence, probability sum of scores is at least 10, is; \(\frac{n(E)}{n(S)}\) = \(\frac{6}{36}\)

= \(\frac{1}{6}\)

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JAMB 1987

In the figure, PQRS is a rectangle. If the shaded area is 72 sq. cm, find h.

• A. 12cm
• B. 10cm
• C. 8cm
• D. 5cm

From the diagram, PQRS is a rectangle Area of shaded part = 72sq.cm But 72 = 3h - 4 + 6h - 4 + 4h = 72 - 16h - 8 = 16h - 72 + 8 =16h = 80 h = \(\frac{80}{16}\)

= 5cm

When two dice are thrown, the sample space is

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),         (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),         (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),         (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),         (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

        (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n (S) = 36

Let event A: Sum of the numbers on uppermost face is 5.

∴ A = {(1, 4), (2, 3), (3, 2), (4 ,1)}

∴ n(A) = 4

∴ P(A) = `("n"("A"))/("n"("S"))`

= `4/36`

= `1/9`.

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When a fair die is tossed twice, the sample sp ace S is given by

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),         (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),         (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),         (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),         (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

        (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n (S) = 36

Let B = event that sum of numbers is at least 8

∴ B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6)}

∴ n(B) = 15

∴ P(B) = `("n"("B"))/("n"("S"))`

= `15/36`

= `5/12`