A fair die is tossed twice what is the probability of get a sum of at least 10.

  • A \(\frac{5}{36}\)
  • B \(\frac{2}{3}\)
  • C \(\frac{5}{18}\)
  • D \(\frac{1}{6\)

The correct answer is D. \(\frac{1}{6\)

\(\begin{array}{c|c}

& 1 & 2 & 3 & 4 & 5 & 6 \\

\hline

1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6 \\ \hline 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6 \\ \hline 3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6 \\ \hline 4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6 \\ 5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6 \\ \hline 6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}\)

From the table above, event space, n(E) = 6

sample space, n(S) = 36

Hence, probability sum of scores is at least 10, is;

\(\frac{n(E)}{n(S)}\)

= \(\frac{6}{36}\)

= \(\frac{1}{6}\)

Previous question Next question