(Method 1) If the ball with number 0 is taken as event A for the first time and the ball with number 0 is taken as event B for the second time, the probability of the event is
P(A∪B)= P(A)+P(B)-P(A∪B)= 1/4+ 1/4- 1/4 * 1/4 = 7/ 16
(Method 2) Consider the negative.
Let the number of two balls taken by event A be different from 0, then the probability obtained is
1-p(a)= 1-(3/4)^2=7/ 16
(Method 3) Classical probability, the basic event is a coded combination of two ball fetching, and the total number of basic events is 4*4.
There are three possibilities for the product of two numbers to be 0:
A: The first ball number is 0, the second ball number is not 0, and the basic event number is 3(C3 1).
B: The number of the first ball is not 0, the second ball is 0, and the number of basic items included is also 3.
C: The number of balls taken twice is 0, and the number of basic events included is 1.
A, B and C are mutually exclusive events.
So the calculated probability is p = p (a)+p (b)+p (c) = (3+3+1)/4 * 4 = 7/16,2, C44C44 is wrong, and it should be1once, so it is. You have to subtract the time of repeated calculation. 1, C44*C44 is four of the four balls. The topic requires you to take two, 1. There are three ways to find the probability when X=0:
(Method 1) If the ball with number 0 is taken as event A for the first time and the ball with number 0 is taken as event B for the second time, the probability of the event is
P(A∪B)= P(A)+P(B)-P(A∪B)= 1/4+ 1/4- 1/4 * 1/4 = 7/ 16
(Method 2) Consider the negative.
Let the number of two balls taken by event A be different from 0, then the probability obtained is
1-p(a)= 1-(3/4)^2=7/ 16
(Method 3) Classical probability, two basic events ... 0, on the probability of 2-3 in the senior high school entrance examination of mathematics.
There are four balls numbered 0, 1, 1 2 in a box. Take out two balls in situ. Let x be the product of the extracted numbers, and find the X distribution table.
When X=0, the probability is1/4+1/4-1/4 *1/4. Why can't C 1 14/C44C44 be used?