In each experiment, the probability of an event occurring is p=0.2, and the probability of an event not occurring is 0.8.
Therefore, the probability that just three families subscribe to two newspapers is (C4,3) × 0.23× 0.8 = 0.0256.
2. At most three companies subscribe to B newspaper. The opposite event is that four families did not subscribe to B newspaper, and one
The probability of family subscribing to B newspaper is 0.6, and the probability of not subscribing to B newspaper is 0.4.
The probability of four families not subscribing to B newspaper is 0.4 4 = 0.064. According to the probability of opposing events,
Rate, the probability that at most three companies subscribe to B newspaper is 1-0.064 = 0.936.
3. I remember that one family designated A as event A, and another family designated B as event B.
It can be seen that p(A)=0.3, p(B)=0.6, and the probability of ordering A and B newspapers at the same time is p(AB)=0.2.
∴ The probability that a family subscribes to A newspaper or B newspaper p (a+b) = p (a)+p (b)-p (ab) = 0.6+0.3-0.2 = 0.7.
The probability that a family does not subscribe to newspapers is1-p (a+b) =1-0.7 = 0.3.
∴ The probability that two families have not subscribed to ab newspaper is (C4,2) × 0.72× 0.32 = 0.2646.
Note: (C4,3) indicates the number of combinations of three of the four elements.
Similarly, (c4,2) represents the number of combinations of two of the four elements.