When a die is thrown twice, there are 36 possible outcomes.
"\\begin{matrix}\n (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\\\\n (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\\\\n (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\\\\n (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\\\\n (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\\\\n (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6)\n\\end{matrix}" There are 20 favorable outcomes
"\\begin{matrix}\n (1,3) & (1,6) & (2,3) & (2,6) & (3,1) & (3,2)\\\\\n(3,3) & (3,4) & (3,5) & (3,6) & (4,3) & (4,6) \\\\\n(5,3) & (5,6) & (6,1) & (6,2) & (6,3) & (6,4) \\\\\n(6,5) & (6,6)\n\\end{matrix}" Find the probability of getting a 3 or 6
"P(3\\ or\\ 6)={20 \\over 36}={5 \\over 9}"
Comments
Leave a comment