Question

I toss an unfair coin 12 times. This coin is 65% likely to show up heads. Calculate the probability of the following.

a. 11 heads:
b. 2 or more heads:
c. 7 heads:
d. 9 tails:
e. 8 or less heads:

Answer 1

a. 0.0368

b. 0.99992131

c. 0.2039

d. 0.0048

e. 0.6533

Explanation:

Let the probability of obtaining a head be p = 65% = 13/20 = 0.65. The probability of not obtaining a head is q = 1 - p = 1 -13/20 = 7/20 = 0.35

Since this is a binomial probability, we use a binomial probability.

a. The probability of obtaining 11 heads is ¹²C₁₁p¹¹q¹ = 12 × (0.65)¹¹(0.35) = 0.0368

b. Probability of 2 or more heads P(x ≥ 2) is

P(x ≥ 2) = 1 - P(x ≤ 1)

Now P(x ≤ 1) = P(0) + P(1)

= ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹

= (0.65)⁰(0.35)¹² + 12(0.65)¹(0.35)¹¹

= 0.000003379 + 0.00007531

= 0.0007869

P(x ≥ 2) = 1 - P(x ≤ 1)

= 1 - 0.00007869

= 0.99992131

c. The probability of obtaining 7 heads is ¹²C₇p⁷q⁵ = 792(0.65)⁷(0.35)⁵ = 0.2039

d. The probability of obtaining 7 heads is ¹²C₉q⁹p³ = 220(0.65)³(0.35)⁹ = 0.0048

e. Probability of 8 heads or less P(x ≤ 8) = ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹ + ¹²C₂p²q¹⁰ + ¹²C₃p³q⁹ + ¹²C₄p⁴q⁸ + ¹²C₅p⁵q⁷ + ¹²C₆p⁶q⁶ + ¹²C₇p⁷q⁵ + ¹²C₈p⁸q⁴

= = ¹²C₀(0.65)⁰(0.35)¹² + ¹²C₁(0.65)¹(0.35)¹¹ + ¹²C₂(0.65)²(0.35)¹⁰ + ¹²C₃(0.65)³(0.35)⁹ + ¹²C₄(0.65)⁴(0.35)⁸ + ¹²C₅(0.65)⁵(0.35)⁷ + ¹²C₆(0.65)⁶(0.35)⁶ + ¹²C₇(0.65)⁷(0.35)⁵ + ¹²C₈(0.65)⁸(0.35)⁴

= 0.000003379 + 0.00007531 + 0.0007692 + 0.004762 + 0.01990 + 0.05912 + 0.1281 + 0.2039 + 0.2367

= 0.6533