HEcom wrote:
Tom is equally likely to roll exactly 1 strike as to roll exactly 2 strikes in those 4 frames.
the probability of Tom rolling exactly 1 strike, is a strike in any one of the rolling session
{4/10, 1, 1, 1} + {1, 4/10, 1, 1} + {1, 1, 4/10, 1} + {1, 1 1, 4/10} = 4 * (4/10) = 16/10
the probability of Tom rolling exactly 2 strike, is a strike in any two of the rolling session
{4/10, 4/10, 1, 1} + {1, 4/10, 4/10, 1} + {1, 1, 4/10, 4/10} + {4/10, 1 1, 4/10} = 4 * (4/10) = 16/10
C is the answer
You got the C as the answer right but the solution is not.
the probability of Tom rolling exactly 1 strike, is a strike in any one of the rolling session is
NOT:
{4/10, 1, 1, 1} + {1, 4/10, 1, 1} + {1, 1, 4/10, 1} + {1, 1 1, 4/10} = 4 * (4/10) = 16/10
it is:
{4/10, 0.6, 0.6, 0.6} + {0.6, 4/10, 0.6, 0.6} + {0.6, 0.6, 4/10, 0.6} + {0.6, 0.6, 0.6, 4/10} = 4 * (4/10) * (0.6)^3 = 4*0.0864
AND
the probability of Tom rolling exactly 2 strike, is a strike in any two of the rolling session is
NOT:
{4/10, 4/10, 1, 1} + {1, 4/10, 4/10, 1} + {1, 1, 4/10, 4/10} + {4/10, 1 1, 4/10} = 4 * (4/10) = 16/10
it is:
{0.4, 0.4, 0.6, 0.6} + {0.4, 0.6, 0.4, 0.6} + {0.4, 0.6, 0.6, 0.4} + {0.6, 0.4, 0.4, 0.6} + {0.6, 0.4, 0.6, 0.4} + {0.6, 0.6, 0.4, 0.4} = 6 * (0.4)^2 * (0.6)^2
Now, 4 * 0.4 * 0.6^3 = 6 * 0.4^2 * 0.6^2
Hence, C is correct.