Uh Probability Question Finding a Probability Question, thanks

Here when triple triggers, double will not be triggered, there is a precedence problem, that is.

p(2 and then 3, or, 2 is not triggered and then 3 is triggered, or, 2 is triggered but 3 is not triggered) = ;

where there is no trigger, that is, 1 times: p1=;

The last trigger is 2x: p2=p (first triggered 2 but then not 3) =;

The last trigger is 3 times: p3 = p (2 and then 3, or, no 2 and then 3) =.

...... proYour second question is disgusting =.=

Points per game:

p（-6）=；

p（0）=；

p（6）=；

p（12）=；

p（18）=；

If you play 10 rounds, it will be the above convolution 10 times and it will come out =.= I talked about ......Do the math yourself ==

If you use the central theorem == An approximate ...... can be calculated then

1) Triple probability: 50%.

Twice the probability of triggering: (1-50%)*85%=

The total chance of triggering multiple XP is 50%+

2) The title seems to be a bit problematic.

After a total of 10 rounds, even if you win all, you only have 60 points, no more than 100 ah...

1) Triggers only double 85%* (1-50%)

Triggers only triple (1-85%)*50%.

Both trigger 85%*50%.

The total probability of triggering multiple XP is .

2) 10 games with more than 100 points, that is, at least 17 games to win, how can 10 rounds be enough? Probability 0%.

Thank you for being responsible, I received a reply, is it convenient to leave qq?

p(a b) = p(a) + p(b)-p (spine oak ab).

Since the inclusion is a and b are mutually exclusive, p(ab)=0

So p(a b) = p(a) + p(b) cherry chain = choose c

I really don't know anything about probability theory, so I can only solve it in the simplest permutations and combinations.

Question: Arrange the balls as: blue balls (b1, b2), red balls (r1, r2), and green balls (g1, g2).

Rearranging the treemap:

Take B1 as an example, the following:

b2，r1，r2，g1，g2

where b2 is below: r1r2g1g2

Below r1: b2r2g1g2

Below r2: b2r1g1g2

below g1: b2r1r2g2

Below g2: b2r1r2g1

There are 4 5 6 = 120 cases.

It can be seen that the coexistence of BG is 4 2 2 + 4 2 2 = 32;

Probability p = 32 120 = 4 15.

(1) The most intuitive solution:

6 balls take out 1 green 2 blue, there are three cases: 1) green-blue-blue, 2) blue-green-blue, 3) blue-blue-green.

1) probability of (2 6) * (2 5) * (1 4).

2) probability of (2 6) * (2 5) * (1 4).

3) probability of (2 6) * (1 5) * (2 4).

Add together = 1 10

2) Calculate all the possibilities of drawing 3 out of 6 balls: c (3 above, 6 below) = 20 possibilities (total possibility, that is, sample size).

There are only 2 possibilities for drawing 1 green (1 above, 2 below) = 2.

Drawing 2 blue is only possible (2 above, 2 below) = 1.

Draw 1 green and draw 2 blue "again" possibilities = 1 green possibility x draw 2 blue possibilities = 2x1 = 2

Probability of drawing 1 green and 2 blue = probability of drawing 1 green and "again" drawing 2 blues divided by total possibility = 2 20 = 1 10

3) This problem belongs to hypergeometric distribution, and the probability formula of hypergeometric distribution = 2x1 20 = 1 10 is directly used

Take three out of 6 balls, which is c63 = 20

There is only one green ball and two blue balls taken out, and there are only 2 cases. So the probability is 1 10

f(x,y)=(1-exp)(1-exp).

The distribution of x f(x) = f(x, + = 1-exp.

The probability that both parts will have a lifetime of more than 100 hours is (1-f(100) 2=exp

This is a conditional probability problem, p=(1% 80%) 99% 10% +1% 80%)=

That is to say, the probability is still very low, so even if it is detected to be malignant, there is no need to worry too much, and the probability that it should become a fact is only.

Solution: p=99% 1 - 90%) 1 - 99%) 80%= A: The probability of a patient suffering from a malignant tumor is.

The first time the black ball is taken, and the second time the white ball is taken, and the number of basic events is c(3,1)*c(2,1).

The first time you take the white ball, the second time you take the black ball, the number of basic events is c(2,1)*c(3,1).

The basic event space is c(5,1)*c(4,1).

The probability of obtaining one black and one white p=[c(3,1)*c(2,1)+c(2,1)*c(3,1)] c(5,1)*c(4,1)=3 5

There is a return to the ground to fetch, the first time to take the black ball, the second time to take the white ball, the number of basic events is c(3,1)*c(2,1).

The first time you take the white ball, the second time you take the black ball, the number of basic events is c(2,1)*c(3,1).

The basic event space is c(5,1)*c(5,1).

The probability of obtaining one black and one white p=[c(3,1)*c(2,1)+c(2,1)*c(3,1)] c(5,1)*c(5,1)=12 25

The first question is two-fifths multiplied by three-fourths plus two-fifths multiplied by one-half.

The second question is two-fifths multiplied by three-fifths multiplied by two.