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Math game questions
one

(1) biased towards Party B.

The probability that the sum is even is calculated as follows:1/2x2/5+1/2x2/5 = 2/5 (both even and odd) has the same probability of being odd.

(2) The probability of being biased towards Party B and even numbers is calculated as follows: 2/5x1/4+3/5x1/2 = 2/5 (the calculation method is the same).

two

( 1) 1/5

(2)2/5 X 1/4 X = 1/ 10

(3) The possibility of finding two yellows and one red is the possibility of leaving one yellow and one red, which means that if you ask this question differently, the answer should be the same.

There are two red balls and three yellow balls in the bag. What is the possibility of finding a yellow ball and a red ball at a time?

Calculation: 2/5x3/4+3/5x2/4=3/5.

three

The original player won easily for the same reason as the first question.