Question

A box contains 3 coins, one coin is fair, one coin is two headed and one coin is weighted, so that the probability of heads appearing is \(\dfrac{1}{3}\). A coin is selected at random and tossed, then the probability that head appears, is
1. \(\dfrac{11}{18}\)
2. \(\dfrac{7}{18}\)
3. \(\dfrac{1}{8}\)
4. \(\dfrac{1}{4}\)

Answer 1

Correct Answer - Option 1 : \(\dfrac{11}{18}\)

Concept:

If there are 2 cases A and B with probability P(A) and P(B) respectively, then

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• P(A ∩ B) = P(A) × P(B)

 

Calculation:

Case A: Selecting a coin 

P(A) = \(\rm 1\over3\)

Case B: Head appears

P(B) = \(\rm 1\over2\) (for fair coin) + 1 (for two headed coin) + \(\rm 1\over3\) (for biased coin) 

⇒ P(B) = \(\rm 11\over 6\)

The probability that the head appears 

P(X) = P(A) × P(B)

⇒ P(X) = \(\boldsymbol {\rm {1\over3}\times{11\over6} = {11\over18}}\)