karthikpandian19 wrote:Andy and Bob play a game in which a computer randomly selects two real numbers between 0 and 10. Andy's score is the sum of the numbers and Bob's score is one more than the product of the numbers. If the person with the higher score wins the game, what is the probability that Bob wins?
A. 18%
B. 19%
C. 68%
D. 82%
E. 90%
Let the two numbers = x and y.
P(Bob wins) = 1 - P(Adam wins).
(*Please see the disclaimer below.)
P(Adam wins):
Adam wins when x + y > xy + 1:
y-1 > xy - x
y-1 > x(y-1)
x(y-1) - (y-1) < 0
(y-1)(x-1) < 0.
The inequality is valid when y<1 and x>1 (yielding negative*positive) or y>1 and x<1 (yielding positive*negative).
Since the total range is from 0 to 10:
P(x<1 and y>1) = 1/10 * 9/10 = 9/100.
P(x>1 and y<1) = 9/10 * 1/10 = 9/100.
Since Adam wins in either case, we add the fractions:
P(Adam wins) = 9/100 + 9/100 = 18/100.
Thus, P(Bob wins) = 1 - 18/100 = 82/100.
The correct answer is
D.
Disclaimer:
If either number = 1, the result will be a draw, since the sum will be equal to one more than the product:
x + 1 = x(1) + 1.
Since there are an infinite number of real numbers between 0 and 10, P(1) is infinitely small.
Thus, the solution above ignores this probability.
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