On a normal standard die one of the 21 dots from any one of the six faces is removed at random with each dot equally likely to be chosen. The die is then rolled. If the probability that the top face has an odd number of dots is ( p / q ) where p and q are in their lowest form, find ( p + q ).

Question

On a normal standard die one of the 21 dots from any one of the six faces is removed at random with each dot equally likely to be chosen. The die is then rolled. If the probability that the top face has an odd number of dots is ( p / q ) where p and q are in their lowest form, find ( p + q ). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

E1: event that the dot is removed from an odd face; P(E1)=(1+3+5/21)=(9/21) E 2: dot is removed from the even face; P ( E 2)=(2+4+6/21)=(12/21) E: die thrown has an odd number of dots on its top face P(E)=P(E ∩ E1)+P(E ∩ E2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/d27532e48e60f9c953f0fa27bf2c2805-.png" /> =P(E1) ⋅ P(E / E1)+P(E2) ⋅ P(E / E2) =((9/21)) ⋅ (2/6)+((12/21)) ⋅ (4/6) =(9/21) ⋅ (1/3)+(12/21) ⋅ (2/3)=(3/21)+(8/21)=(11/21)=(p/q) ⇒ p+q=32

Q. On a normal standard die one of the 21 dots from any one of the six faces is removed at random with each dot equally likely to be chosen. The die is then rolled. If the probability that the top face has an odd number of dots is qp​ where p and q are in their lowest form, find (p+q).