0 votes 0 votes A person throwing a dice and when even have come he got $2$ points and when odd have come he got $1$ points.What is the probability when he got exactly $6$ points? Probability engineering-mathematics probability + – Lakshman Bhaiya asked Nov 2, 2018 Lakshman Bhaiya 266 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Lakshman Bhaiya commented Nov 2, 2018 i edited by Lakshman Bhaiya Nov 2, 2018 reply Follow Share I make the case like this $P(6)=P(E)P(E)P(E)+P(E)P(E)P(O)P(O)+P(E)P(O)P(O)P(O)P(O)+P(O)P(O)P(O)P(O)P(O)P(O)$ for $P(E)=\frac{3}{6}=\frac{1}{2}$ and $P(O)=\frac{3}{6}=\frac{1}{2}$ So,$P(6)=\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}+\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}$ $P(6)=\frac{1}{2^{3}}+\frac{1}{2^{4}}+\frac{1}{2^{5}}+\frac{1}{2^{6}}$ $P(6)=\frac{14}{64}$ but answer is $\frac{43}{64}??$ 0 votes 0 votes Mk Utkarsh commented Nov 2, 2018 reply Follow Share $3$ consecutive even numbers = $(\frac{1}{2})^3$ 2 Even and 2 Odd numbers can be arranged in $\frac{4!}{2! 2!} = 6$ = $6 \times (\frac{1}{2})^4$ similarly count for other cases like 4 odd 1 even, all odd ,etc 0 votes 0 votes Lakshman Bhaiya commented Nov 2, 2018 reply Follow Share @Mk Utkarsh In your explanation i got $\frac{45}{64}?$ 0 votes 0 votes Lakshman Bhaiya commented Nov 2, 2018 reply Follow Share Thanks now I got the answer:) 0 votes 0 votes Please log in or register to add a comment.