## Xinmin Secondary 2010 Prelim Paper I Q24 Solution (Challenging/Difficult Probability O Level Question)

A bag A contains 9 black balls, 6 white balls and 3 red balls. A bag B contains 6 black balls, 2 white balls and 4 green balls. Ali takes out 1 ball from each bag randomly. When Ali takes out 1 ball from one bag, he will put it into the other bag and then takes out one ball from that bag. Find the probability that

(a) the ball is black from bag A, followed by white from bag B,
(b) both the balls are white in colour,
(c) the ball is black or white from bag B, followed by red from bag A,
(d) both the balls are of different colours,
(e) both the balls are not black or white in colours.

Solution:

(a) $\displaystyle\frac{9}{18}\times\frac{2}{13}=\frac{1}{13}$

(b) Probability of white ball from bag A, followed by white ball from bag B=$\displaystyle=\frac{1}{2}\times\frac{6}{18}\times\frac{3}{13}=\frac{1}{26}$

Probability of white from B, followed by white from A=$\displaystyle=\frac{1}{2}\times\frac{2}{12}\times\frac{7}{19}=\frac{7}{228}$

Total prob=$\displaystyle\frac{205}{2964}$

(c) Prob. of ball is black or white from bag B=$\displaystyle\frac{6}{12}+\frac{2}{12}=\frac{8}{12}$

$\displaystyle\frac{8}{12}\times\frac{3}{19}=\frac{2}{19}$

(d) Prob of both red = P(red from A, followed by red from B)=$\displaystyle\frac{1}{2}\times\frac{3}{18}\times\frac{1}{13}=\frac{1}{156}$

P(both green)=P(green from B, followed by green from A)=$\displaystyle\frac{1}{2}\times\frac{4}{12}\times\frac{1}{19}=\frac{1}{114}$

P(both black)=P(black from A, followed by black from B)+P(black from B, followed by black from A)=$\displaystyle\frac{1}{2}\times\frac{9}{18}\times\frac{7}{13}+\frac{1}{2}\times\frac{6}{12}\times\frac{10}{19}=\frac{263}{988}$

P(both white)=$\displaystyle\frac{205}{2964}$ (from part b)

$\displaystyle 1-\frac{1}{156}-\frac{1}{114}-\frac{263}{988}-\frac{205}{2964}=\frac{1925}{2964}$

(e)

P(neither black nor white from A, followed by neither black nor white from B)=$\displaystyle\frac{1}{2}\times\frac{3}{18}\times\frac{5}{13}=\frac{5}{156}$

P(neither black nor white from B, followed by neither black nor white from A)=$\displaystyle\frac{1}{2}\times\frac{4}{12}\times\frac{4}{19}=\frac{2}{57}$

$\displaystyle\frac{5}{156}+\frac{2}{57}=\frac{199}{2964}$

## MI H2 Maths Prelim Solutions 2010 Paper 2 Q7 (P&C)

MI H2 Maths Prelim 2010 Paper 2 Q7 (P&C)

Question:
7) Two families are invited to a party. The first family consists of a man and both his parents while the second family consists of a woman and both her parents. The two families sit at a round table with two other men and two other women.
Find the number of possible arrangements if
(i) there is no restriction, [1]
(ii) the men and women are seated alternately, [2]
(iii) members of the same family are seated together and the two other women must be seated separately, [3]
(iv) members of the same family are seated together and the seats are numbered. [2]

Solution:

(i) (10-1)!=9!=362880

(ii) First fix the men’s sitting arrangement: (5-1)!

Then the remaining five women’s total number of arrangements are: 5!

Total=4! x 5!=2880

(iii) Fix the 2 families (as a group) and the 2 men: (4-1)! x 3! x 3!

(3! to permute each family)

By drawing a diagram, the two women have 4 slots to choose from, where order matters: $^4 P_2$

Total = $(4-1)! \times 3! \times 3! \times ^4 P_2 = 2592$

(iv)

We first find the required number of ways by treating the seats as unnumbered: $(6-1)!\times 3!\times 3! =4320$

Since the seats are numbered, there are 10 choices for the point of reference, thus no. of ways = $4320 \times 10 =43200$