Probability

Miscellaneous Exercise Solutions
Key Concepts:
Combinations: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$
Complementary Probability: $P(A’) = 1 – P(A)$
Probability of Union: $P(A \cup B) = P(A) + P(B) – P(A \cap B)$
Q1-Q2
1. Box: 10 Red, 20 Blue, 30 Green. Draw 5.
Total Total marbles = $10+20+30=60$. Total outcomes = $\binom{60}{5}$.
(i) All Blue Favorable: Choose 5 from 20 Blue. $\binom{20}{5}$.
Prob = $\frac{\binom{20}{5}}{\binom{60}{5}}$.
(ii) At least 1 Green $1 – P(\text{No Green})$. No Green means choosing from 30 non-green (10R + 20B).
Prob = $1 – \frac{\binom{30}{5}}{\binom{60}{5}}$.
2. Draw 4 cards. Prob of 3 Diamonds and 1 Spade.
Method Total outcomes = $\binom{52}{4}$.
Favorable: Select 3 from 13 Diamonds ($\binom{13}{3}$) AND 1 from 13 Spades ($\binom{13}{1}$).
$P = \frac{\binom{13}{3} \times \binom{13}{1}}{\binom{52}{4}}$
Q3-Q4
3. Die faces: two ‘1’, three ‘2’, one ‘3’.
Sample Space Total faces = 6. Frequencies: $n(1)=2, n(2)=3, n(3)=1$.
(i) P(2) $\frac{3}{6} = \frac{1}{2}$.
(ii) P(1 or 3) $P(1) + P(3) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$.
(iii) P(not 3) $1 – P(3) = 1 – \frac{1}{6} = \frac{5}{6}$.
4. Lottery: 10,000 tickets, 10 prizes. Prob of NOT winning.
Tickets Total = 10,000. Winners = 10. Losers = 9,990.
(a) 1 Ticket $P(\text{Loss}) = \frac{9990}{10000} = \frac{999}{1000}$.
(b) 2 Tickets Both must be losers. $\frac{\binom{9990}{2}}{\binom{10000}{2}}$.
(c) 10 Tickets All must be losers. $\frac{\binom{9990}{10}}{\binom{10000}{10}}$.
Q5-Q6
5. 100 Students (Sections of 40 and 60). You and Friend.
Total Pairs Total ways to select 2 students from 100 = $\binom{100}{2}$.
(a) Same Sec Both in 40 ($\binom{40}{2}$) OR Both in 60 ($\binom{60}{2}$).
Prob = $\frac{\binom{40}{2} + \binom{60}{2}}{\binom{100}{2}} = \frac{780 + 1770}{4950} = \frac{2550}{4950} = \frac{17}{33}$.
(b) Diff Sec $1 – P(\text{Same}) = 1 – \frac{17}{33} = \frac{16}{33}$.
6. 3 Letters, 3 Envelopes. Prob of at least one in proper envelope.
Logic Total arrangements = $3! = 6$.
Event E: At least one correct. $E’ = \text{None correct}$ (Derangement of 3).
Calc Derangements of 3 items = 2 ways (e.g., if correct is 1,2,3 -> then 2,3,1 and 3,1,2 are wrong).
$P(\text{None}) = \frac{2}{6} = \frac{1}{3}$.
Result $P(\text{At least one}) = 1 – \frac{1}{3} = \frac{2}{3}$.
Q7
7. $P(A)=0.54, P(B)=0.69, P(A \cap B)=0.35$.
(i) Union $P(A \cup B) = 0.54 + 0.69 – 0.35 = 0.88$.
(ii) A’ $\cap$ B’ $P((A \cup B)’) = 1 – P(A \cup B) = 1 – 0.88 = 0.12$.
(iii) A $\cap$ B’ $P(A) – P(A \cap B) = 0.54 – 0.35 = 0.19$.
(iv) B $\cap$ A’ $P(B) – P(A \cap B) = 0.69 – 0.35 = 0.34$.
Q8
8. Select 1 person. Prob(Male or Age > 35).
Data Total = 5.
Males: Harish, Rohan, Salim (3).
Age > 35: Sheetal (46), Salim (41) (2).
Intersection Male AND Age > 35: Salim (1 person).
Formula $P(M \cup >35) = P(M) + P(>35) – P(M \cap >35) = \frac{3}{5} + \frac{2}{5} – \frac{1}{5} = \frac{4}{5}$.
4/5
Q9
9. 4-digit numbers > 5000 using {0, 1, 3, 5, 7}. Prob(Divisible by 5).
Case (i) Repetition Allowed
Total > 5000: Thousands (5, 7) $\times$ 5 $\times$ 5 $\times$ 5 = 250.
Exclude 5000 (starts 5, rest 0s, allowed) since strict >. Total = 249.
Favorable (Div 5): Ends in 0 or 5.
Start {5, 7} (2 options) $\times$ Any (5) $\times$ Any (5) $\times$ End {0, 5} (2 options) = 100.
Exclude 5000 (It ends in 0, so it’s in this set). Fav = 99.
Prob = $99/249 = 33/83$.
Case (ii) No Repetition
Total > 5000: Starts with 5 or 7. $2 \times 4 \times 3 \times 2 = 48$.
Favorable (End 0 or 5):
1. Start 5 (must end 0): $1 \times 3 \times 2 \times 1 = 6$.
2. Start 7 (end 0): $1 \times 3 \times 2 \times 1 = 6$.
3. Start 7 (end 5): $1 \times 3 \times 2 \times 1 = 6$.
Total Fav = 18.
Prob = $18/48 = 3/8$.
Q10
10. Number lock: 4 wheels (0-9). Sequence with no repeats. Prob of opening.
Total Seqs Permutation of 4 digits from 10. $10 \times 9 \times 8 \times 7 = 5040$.
Correct Only 1 correct sequence.
1/5040
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