WAEC 2010 · Paper 2 · Q5

Probability

(a)
Write out the elements of $M$ and $N$ .
Show the answer
$M = \{(4,6), (5,5), (6,4)\}$ ; $N = \{(1,4), (4,1), (2,5), (5,2), (3,6), (6,3)\}$
(b)
Find the probability of $M$ or $N$ .
(c)
Are $M$ and $N$ mutually exclusive? Give reasons.
Show the answer
Yes: they have no outcome in common, $M \cap N = \varnothing$ .

Worked solution (try it first)

(a)

Write each outcome as (first die, second die).
A sum of 10 comes from $M = \{(4,6), (5,5), (6,4)\}$ .
A difference of 3 comes from $N = \{(1,4), (4,1), (2,5), (5,2), (3,6), (6,3)\}$ .

(b)

There are $6 \times 6 = 36$ equally likely outcomes, so $P(M) = \frac{3}{36} = \frac{1}{12}$ and $P(N) = \frac{6}{36} = \frac16$ .
No outcome is in both, so add: $P(M \text{ or } N) = \frac{1}{12} + \frac{2}{12}$
$= \frac{3}{12}$
$= \frac14$ .

(c)

Yes, $M$ and $N$ are mutually exclusive: no outcome is in both lists ( $M \cap N = \varnothing$ ), so they cannot happen together.

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