Finding the Probability of Serendipity Full Text 5

As can be seen from the inscription, ball No. 1234 is white and ball No. 567 is red, so the ball taken out is not a red ball with an odd number, which means that the ball taken out is white and odd-numbered, so there are only two balls on No. 13 1 c[27]=1 21

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There are two difficulties in this topic, mainly plane analytic geometry. In the geometric generalization, just pay attention to the range of basic events.

For reference, please smile.

Probability that both ab events do not occur at the same time =

The probability that two events do not happen at the same time = the probability that event A does not happen * the probability that event B does not happen = the probability that the event does not occur, then the probability that event A does not happen = , so P(A) = choose C

Choose B, because they are independent of each other, so there is no B in A, and A does not have in B, so choose B

Let Event A be a forgery, and the remaining 4 paintings are fakes.

Let Event B be a random sample to be examined, and the result will be a fake.

The probability is prob(a|).b) =prob(ab) prob(b)prob(ab) =prob(all 5 paintings are fakes) =prob(b) =prob(the drawn is a fake) *prob (1 painting is a fake) +prob (the drawn is a fake) *prob (2 paintings are fakes) +prob (the drawn is a fake) * prob (5 paintings are fakes) = 1 5 * =

prob(a|b) =prob(ab) /prob(b) = =

1 All 1The odd-numbered balls placed in the box numbered 1 can be , so the probability is 1 2;

2.After the odd ball is placed in the box number 1, there is one odd ball left, and three boxes, then the probability of that odd ball being put into each box is 1 3,3So the probability value = 1 2 1 3 = 1 6

There are a total of 44 types that are 24 types.

There are only 4 situations that are eligible.

So the probability is 1 6

The odd number of balls into the odd number of the four cases: 1 1, 1 3, 3 1, 3 3, and the total number of cases are: a44 = 4 * 3 * 2 * 1 = 24 kinds.

The probability value of an odd-numbered ball being placed in an odd-numbered box: p 4 24 = 1 6

The probability of the first success is 1 2, and the probability of the second success is 1 2*1 2+3 4*1 2=5 8 Let the probability of the nth success be p(n), then p(n)=1 2*p(n-1)+3 4*[1-p

1. The probability of winning the lottery once is p1=1 10

The probability of winning the lottery in one draw is: p2 = 1 10

The probability of neither A nor B winning the lottery is: p=(1-1 10)(1-1 10) 2=729 1000

2、.The probability of at least one of A and B winning the second prize is: 1-p=271 1000

1. A=A wins the lottery P(A)=

b = B wins: p(b) =

The probability that at least one of A and B will win the lottery is: p(a b)=The probability that both A and B have not won the lottery is: p=1-p(a b)=2, and the probability of winning the second prize in A is:

p(a)=,c=second prize in B: p(c)=.The probability that at least one of A and B winning the second prize is p(ab)=