Question 4: How to calculate the probability? How to solve the fourth problem of probability

This is a binomial distribution, each sampling is exactly the same and independent because there is a return, assuming that the probability of drawing two reds in each sampling is p, and the probability of drawing non-two reds (red and white or two white) is 1-p, then according to the formula of the binomial distribution, the probability of three sampling exactly two times satisfied is p*p*(1-p)*c(3,2), where c(3,2)=3.

where p refers to exactly two red balls drawn in a single draw, which requires 1The first one drawn is a red ball (2 4), 2The second red ball (1 3) is drawn from the remaining three balls, using a step-by-step count, multiplying the probabilities, p=1 6

Bringing p into the above formula, we can get that the probability of the problem event is (1 6) (1 6) (5 6) * 3 = 5 72

The probability of catching the red ball twice is 1 2x1 2=1 4 and catching three times, and the probability of this process belongs to the binomial distribution.

b(3,1/4)

Then the probability of both red balls being red:

c(3,2)*(1/4)^2*(1-1/4)

The probability of taking out a red ball each time is 1 2, and the probability of being a red ball three times in a row is 1 2 * 1 2 * 1 2 = 1 8, and exactly two times are red balls: the first and second, the first and third, the second and the third --" 1 8 + 1 8 + 1 8 + 1 8 = 3 8

The probability that both are red balls is 50%.

Because of p(a).

So p(a inverse).

and p(b|A reverse).

According to the probability multiplication formula p (ba inverse).

p(b|a-inverse) p(a-inverse).

And according to the formula of probability addition, p(b).

p(ab)p(ainverseb)p(ab).

So the answer is p(ab).

Note that x and y are the transmitted and received signals, then.

p(x=1|y=1) = p(y=1,x=1)/[p(y=1|x=1)*p(x=1)+p(y=1|x=0)*p(x=0)] =

1. The landlord must first know that p(aubuc)=p(a)+p(b)+p(c)-p(ab)-p(bc)-p(ac)+p(abc)If you don't understand this equation, you can use Wayne's diagram to draw the diagram where three circles intersect each other; Or take a look at the proof:

Since p(a b) = p(a) + p(b)-p(ab).

Then: p((a b) c) = p(a b) + p(c)-p((a b)c).

p(a)+p(b)-p(ab)+p(c)-p(ac∪bc)

p(a)+p(b)+p(c)-p(ab)-p(ac)-p(bc)+p(ac∩bc)

p(a)+p(b)+p(c)-p(ab)-p(ac)-p(bc)+p(abc)

2. The three events are independent of each other, so the probability of the intersection of the two events is equal to the product of the probability of each event. Assuming p(a)=p(b)=p(c)=x, then p(ab)=p(bc)=p(ac)=x 2, and abc= ,p(abc)=0 thus the unary quadratic equation 3x-3x 2=9 16

3. The equation has two roots, one and one, according to p(a)=p(b)=p(c), know that it is the answer.

Thus p(a)=1 4

Let p(a) = x, then: p(b) = p(c) = x Because the two are independent, so: p(ab) = p(a)p(b) = x 2, p(bc) = p(ca) = x 2

p(a∪b∪c) = p(a) +p(b) +p(c) -p(ab) -p(bc) -p(ca) +p(abc) = 3x - 3x^2 = 9/16

That is: 16x 2 - 16x + 3 = 0, which is: (4x-1)(4x-3) = 0

So: x = 1 4 or 3 4

And since the question is given: x < 1 2, so: x = 1 4