Chapter 13: Probability
Key Formulas:
• P(A|B) = P(A ∩ B)/P(B)
• Bayes' Theorem: P(E₁|A) = P(A|E₁)P(E₁)/Σ P(A|E₁)P(E₁)
• P(X = r) = nCₓ pr qn−r (Binomial)
Q1. A bag contains 4 red and 4 black balls. Another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn. If the ball is red, find the probability that it was drawn from the first bag.
Solution
Let E₁ = first bag, E₂ = second bag, A = red ball
P(E₁) = P(E₂) = 1/2
P(A|E₁) = 4/8 = 1/2
P(A|E₂) = 2/8 = 1/4
P(E₁|A) = P(A|E₁)P(E₁)/[P(A|E₁)P(E₁) + P(A|E₂)P(E₂)]
= (1/2 · 1/2)/[(1/2 · 1/2) + (1/4 · 1/2)]
= (1/4)/(1/4 + 1/8) = (1/4)/(3/8) = 2/3
Q2. A coin is tossed 3 times. What is the probability of getting exactly 2 heads?
Solution
n = 3, p = 1/2, q = 1/2, r = 2
P(X = 2) = 3C₌ (1/2)² (1/2)¹
= 3 · 1/4 · 1/2 = 3/8
Q3. An urn contains 5 red and 5 black balls. Two balls are drawn at random one by one without replacement. Find the probability that both are red.
Solution
P(both red) = P(R₁) · P(R₂|R₁)
= 5/10 · 4/9 = 20/90 = 2/9
Q4. A and B throw a pair of dice alternately. A wins if he gets a sum of 6, B wins if he gets a sum of 7. If A starts the game, find the probability that A wins.
Solution
P(A gets sum 6) = 5/36, P(not) = 31/36
P(B gets sum 7) = 6/36 = 1/6, P(not) = 30/36 = 5/6
P(A wins) = P(A on 1st) + P(A on 3rd) + P(A on 5th) + ...
= 5/36 + (31/36)(5/6)(5/36) + (31/36)²(5/6)²(5/36) + ...
This is a G.P. with a = 5/36, r = (31/36)(5/6) = 155/216
P(A wins) = (5/36)/(1 − 155/216) = (5/36)/(61/216) = (5/36) · (216/61) = 30/61
Q5. Find the probability distribution of the number of heads in 2 tosses of a coin.
Solution
X can take values 0, 1, 2.
P(X=0) = P(TT) = 1/4
P(X=1) = P(HT) + P(TH) = 1/4 + 1/4 = 1/2
P(X=2) = P(HH) = 1/4
Distribution:
X | 0 1 2
P(X) | 1/4 1/2 1/4
Mean = 0(1/4) + 1(1/2) + 2(1/4) = 1
Q6. Two cards are drawn successively without replacement from a deck of 52 cards. Find the probability that both are aces.
Solution
P(both aces) = P(first ace) · P(second ace | first ace)
= 4/52 · 3/51 = 1/13 · 1/17 = 1/221