Probability

The man has 6 relatives and $10 - 6 = 4$ non-relatives. To invite 5 guests with exactly 2 relatives, choose 2 from the 6 relatives and the other 3 from the 4 non-relatives: $\binom62\times\binom43 = 15\times 4 = 60$ He can invite the guests in 60 ways.

Let $P(R) = \tfrac14$ and $P(S) = \tfrac15$, so $P(\bar R) = \tfrac34$ and $P(\bar S) = \tfrac45$. The two selections are independent.

Both selected: $P(R)\,P(S) = \frac14\times\frac15 = \frac{1}{20}$

Only one selected: $P(R)P(\bar S) + P(\bar R)P(S) = \frac14\times\frac45 + \frac34\times\frac15 = \frac{4}{20}+\frac{3}{20} = \frac{7}{20}$

None selected: $P(\bar R)\,P(\bar S) = \frac34\times\frac45 = \frac{12}{20} = \frac35$

Check: $\tfrac1{20}+\tfrac7{20}+\tfrac{12}{20} = 1$.

(c) $\dfrac12$

The prime numbers on a die are 2, 3 and 5, so there are 3 favourable outcomes out of 6. Hence $P=\dfrac36=\dfrac12$.

Total number of possible outcomes when a card is drawn from a well-shuffled deck:

$n(S)=52$

(i) A king or a queen

There are 4 kings and 4 queens.

$n(A)=4+4=8$

$P(\text{king or queen})=\frac{8}{52}=\frac{2}{13}$

(ii) A red card or a face card

Red cards = 26
Face cards = 12
Red face cards = 6

Using the addition theorem:

$P(\text{red or face})=\frac{26+12-6}{52}=\frac{32}{52}=\frac{8}{13}$

(iii) Neither an ace nor a jack

Aces = 4
Jacks = 4
Total ace or jack cards = 8

So, cards that are neither ace nor jack:

$52-8=44$

$P(\text{neither ace nor jack})=\frac{44}{52}=\frac{11}{13}$

Answers:

$(i)\ \frac{2}{13}\qquad (ii)\ \frac{8}{13}\qquad (iii)\ \frac{11}{13}$

Let

• $n(M)$ = number who passed Mathematics = 25
• $n(E)$ = number who passed English = 20
• $n(M \cap E)$ = number who passed both = 10
• Total students $n(S)$ = 40

(i) Passed at least one of the two subjects

Students who passed at least one subject:

So,

(ii) Passed neither Mathematics nor English

Students who passed neither subject:

So,

Answer:

(i) Tree diagram

Total balls initially $=5+4=9$.

                         ┌── R (4/8) ── RR : 5/18
              ┌── R (5/9)
              │          └── B (4/8) ── RB : 5/18
Start ────────┤
              │          ┌── R (5/8) ── BR : 5/18
              └── B (4/9)
                         └── B (3/8) ── BB : 1/6

So the possible outcomes are:

$RR,\quad RB,\quad BR,\quad BB$

(ii) Probability that both balls are red

After drawing one red ball, the bag contains:

$4R,\ 4B,\quad \text{total}=8$

So,

$P(RR)=P(R_1)\times P(R_2/R_1)$

$=\frac{5}{9}\times \frac{4}{8}$

$=\frac{5}{9}\times \frac{1}{2}$

$=\frac{5}{18}$

$\boxed{P(\text{both red})=\frac{5}{18}}$

(iii) Probability that the first ball is red and the second ball is black

$P(RB)=P(R_1)\times P(B_2/R_1)$

$=\frac{5}{9}\times \frac{4}{8}$

$=\frac{5}{18}$

$\boxed{P(\text{first red, second black})=\frac{5}{18}}$

(iv) Probability that at least one ball is black

At least one black means the outcomes are:

$RB,\quad BR,\quad BB$

Alternatively,

$P(\text{at least one black})=1-P(RR)$

$=1-\frac{5}{18}$

$=\frac{18-5}{18}$

$=\frac{13}{18}$

$\boxed{P(\text{at least one black})=\frac{13}{18}}$