<p>A child’s toy has $36$ slots, numbered from $1$ to $36$. A child puts a ball into the toy, the ball falls into one of the $36$ slots and the child’s score is the number of that slot. The ball is equally likely to fall into any one of the slots.</p>
<p>Sadiq is investigating four possible events, $A$, $B$, $C$, and $D$, which are defined as follows.</p>
<p style="margin-left: 3rem">$A$:	The score is odd.</p>
<p style="margin-left: 3rem">$B$:	The score is even.</p>
<p style="margin-left: 3rem">$C$:	The score is a multiple of $3$.</p>
<p style="margin-left: 3rem">$D$:	The score is a multiple of $6$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-paragraph">State which pairs of the events, if any, are mutually exclusive.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Show that $A$ and $C$ are independent events, and state another pair of independent events.</div><p style="text-align: right;">[2]</p></li></ol></li></ol>
<p>Sadiq notices that a ball has become stuck in the slot labelled $36$, and so balls put into the toy are now falling, with equal probability, into one of only $35$ slots, and the score can only be from $1$ to $35$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-paragraph">State which pairs of the four events, if any, are now mutually exclusive.</div><p style="text-align: right;">[1]</p></li><li><div class="list-item-paragraph">Determine whether $A$ and $C$ are now independent events.</div><p style="text-align: right;">[1]</p></li></ol></li></ol>

A child’s toy has $36$ slots, numbered from $1$ to $36$. A child puts a ball into the toy, the ball falls into one of the $36$ slots and the child’s score is the number of that slot. The ball is equally likely to fall into any one of the slots.

Sadiq is investigating four possible events, $A$, $B$, $C$, and $D$, which are defined as follows.

Sadiq notices that a ball has become stuck in the slot labelled $36$, and so balls put into the toy are now falling, with equal probability, into one of only $35$ slots, and the score can only be from $1$ to $35$.

Probability

H2 Math

H1 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

N2023 P2 Q5

$A = \{1, 3, 5, 7, 9, \ldots, 35\}$ (odd) $\Rightarrow n(A) = \dfrac{36}{2} = 18$

$A' = B = \{2, 4, 6, 8, \ldots, 36\}$ (even) $\Rightarrow n(B) = 18$

$C = \{3, 6, 9, 12, \ldots, 36\}$ (multiples of 3) $\Rightarrow n(C) = \dfrac{36}{3} = 12$

$D = \{6, 12, 18, 24, 30, 36\}$ (multiples of 6) $\Rightarrow n(D) = \dfrac{36}{6} = 6$

odd × odd = odd: LHS = $(2k+1)(2n+1) = 4kn + 2k + 2n + 1 = 2(2kn + k + n) + 1$ = RHS

odd × even = even: LHS = $(2k+1)(2n) = 2[n(2k+1)]$ which is even

odd + even = odd: LHS = $(2k+1) + 2n = 2(k+n) + 1$ which is odd

$\therefore$ A and B are mutually exclusive.

$P(A) = \dfrac{18}{36} = \dfrac{1}{2}$, $P(C) = \dfrac{12}{36} = \dfrac{1}{3}$

$P(A \cap C) = \dfrac{n(A \cap C)}{n(S)} = \dfrac{6}{36} = \dfrac{1}{6}$

$P(A) \cdot P(C) = \dfrac{1}{2} \cdot \dfrac{1}{3} = \dfrac{1}{6} = P(A \cap C)$

$\therefore$ A & C are independent events.

$\therefore$ $A' \& C$ are independent, as well as $A' \& C'$, $A \& C'$

$n(A) = 18$, $n(B) = 17$, $n(C) = 11$, $n(D) = 5$

(i) A & B are mutually exclusive. A & D are mutually exclusive.

$P(A) = \dfrac{18}{35}$, $P(C) = \dfrac{11}{35}$

$P(A) \cdot P(C) = \dfrac{18}{35} \times \dfrac{11}{35} = \dfrac{198}{1225}$

Since $P(A)P(C) \neq P(A \cap C)$, A & C are no longer independent events.

N2023 P2 Q5

N2023 P2 Q5

Video Solution: