A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let a = P ( X = 3 ) , b = P ( X ≥ 3 ) and c = P ( X ≥ 6 ∣ X > 3 ) . Then b + c a is equal to

Step-by-step solution

P getting 6 = 1 6 , P not getting 6 = 5 6 ⇒ a = P ( X = 3 ) ⇒ a = 5 6 × 5 6 × 1 6 ⇒ a = 25 216 . . . i ⇒ b = P ( X ≥ 3 ) ⇒ b = 5 6 × 5 6 × 1 6 + 5 6 3 · 1 6 + 5 6 4 · 1 6 + . . . We know that, S ∞ = a + ar + ar 2 + . . . ∞ = a 1 - r ⇒ b = 25 216 1 - 5 6 ⇒ b = 25 216 × 6 1 ⇒ b = 25 36 . . . i i Now, P ( X ≥ 6 ) = 5 6 5 · 1 6 + 5 6 6 · 1 6 + . . . ⇒ P ( X ≥ 6 ) = 5 6 5 · 1 6 1 - 5 6 ⇒ P ( X ≥ 6 ) = 5 6 5 ⇒ c = P X ≥ 6 X > 3 = P X ≥ 6 ∩ X > 3 P X > 3 = P X ≥ 6 P X > 3 = 5 6 5 5 6 3 ⇒ c = 25 36 . . . i i i Using i , i