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, then the value of (p+q)/4 is

Arely Savage

Arely Savage

Answered question

2022-11-26

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, then the value of (p+q)/4 is

Answer & Explanation

Aldo Rios

Aldo Rios

Beginner2022-11-27Added 8 answers

E 1 : Dot removed from odd face.
P ( E 1 ) = 9 21
E 2 : Dot removed from even face.
P ( E 2 ) = 12 21
E: Die shows odd numbers of dots.
P ( E ) = P ( E E 1 ) + P ( E E 2 )
= P ( E 1 ) P ( E E 1 ) + P ( E 2 ) P ( E E 2 )
= 9 21 × 2 6 + 12 21 × 4 6
= 11 21 = p q
Now, 11 + 21 4 = 8

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