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2020
P2 Q8
Topical
P&C
20 P2 Q8
Topical
Probability
20 P2 Q8
2020 H2 Mathematics Paper 2 Question 8
Permutations and Combinations (P&C)
Probability
Answers
(ii)
r
=
6
{r=6}
r
=
6
Full solutions
(i)
P
(
R
=
1
)
=
(
17
1
)
(
11
11
)
(
28
12
)
=
1
1789515
\begin{align*} \textrm{P}\left(R=1\right) & = \frac{{17 \choose 1}{{11 \choose 11}}}{{28 \choose 12}} \\ & = \frac{1}{1789515} \end{align*}
P
(
R
=
1
)
=
(
12
28
)
(
1
17
)
(
11
11
)
=
1789515
1
P
(
R
=
2
)
=
(
17
2
)
(
11
10
)
(
28
12
)
=
88
1789515
\begin{align*} \textrm{P}\left(R=2\right) & = \frac{{17 \choose 2}{{11 \choose 10}}}{{28 \choose 12}} \\ & = \frac{88}{1789515} \end{align*}
P
(
R
=
2
)
=
(
12
28
)
(
2
17
)
(
10
11
)
=
1789515
88
Hence
P
(
R
=
1
)
<
P
(
R
=
2
)
■
{\textrm{P}\left(R=1\right) < \textrm{P}\left(R=2\right) \; \blacksquare}
P
(
R
=
1
)
<
P
(
R
=
2
)
■
(ii)
P
(
R
=
4
)
=
15
P
(
R
=
3
)
(
17
+
r
4
)
(
11
8
)
(
28
+
r
12
)
=
15
(
17
+
r
3
)
(
11
9
)
(
28
+
r
12
)
(
17
+
r
4
)
(
11
8
)
=
15
(
17
+
r
3
)
(
11
9
)
(
17
+
r
)
!
4
!
(
13
+
r
)
!
165
=
15
⋅
(
17
+
r
)
!
3
!
(
14
+
r
)
!
⋅
55
14
+
r
=
15
×
55
×
4
!
3
!
×
165
14
+
r
=
20
r
=
6
■
\begin{align*} \textrm{P}\left(R=4\right) &= 15 \; \textrm{P}\left(R=3\right) \\ \frac{{17+r \choose 4}{11 \choose 8}}{28+r \choose 12} &= \frac{15{17+r \choose 3}{11 \choose 9}}{28+r \choose 12} \\ {17+r \choose 4}{11 \choose 8} &= 15{17+r \choose 3}{11 \choose 9} \\ \frac{(17+r)!}{4!(13+r)!} 165 &= 15\cdot\frac{(17+r)!}{3!(14+r)!}\cdot 55 \\ 14+r &= \frac{15\times55\times4!}{3!\times165} \\ 14+r &= 20 \\ r &= 6 \; \blacksquare \end{align*}
P
(
R
=
4
)
(
12
28
+
r
)
(
4
17
+
r
)
(
8
11
)
(
4
17
+
r
)
(
8
11
)
4
!
(
13
+
r
)!
(
17
+
r
)!
165
14
+
r
14
+
r
r
=
15
P
(
R
=
3
)
=
(
12
28
+
r
)
15
(
3
17
+
r
)
(
9
11
)
=
15
(
3
17
+
r
)
(
9
11
)
=
15
⋅
3
!
(
14
+
r
)!
(
17
+
r
)!
⋅
55
=
3
!
×
165
15
×
55
×
4
!
=
20
=
6
■
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