At most 1 person is extremely happy, of which 1 person is extremely happy and 0 person is extremely happy.
Then p (a) = p (A0)+p (a 1) = c (3, 12)/c (3, 16)+c (1 4) c (2,/kloc).
(2) 1 person is randomly selected from the sample data of 16 people, and the probability of "extremely happy" people is 4/ 16= 1/4.
The possible values of ξ are 0, 1, 2, 3.
p(ξ=0)=c(0,3)( 1/4)^0( 1- 1/4)^3=27/64
p(ξ= 1)=c( 1,3)( 1/4)^ 1( 1- 1/4)^2=27/64
p(ξ=2)=c(2,3)( 1/4)^2( 1- 1/4)^ 1=9/64
p(ξ=3)=c(3,3)( 1/4)^3( 1- 1/4)^0= 1/64
Then the distribution list of ξ is: (omitted)
So e = 0 * 27/64+1* 27/64+2 * 9/64+3 *1/64 = 0.75.
Another solution: the possible values of ξ are 0, 1, 2, 3.
Then ξ ~ b (3, 1/4), p (ξ = k) = c (k, 3) (1/4) k (3/4) (3-k).
∴Eξ=3* 1/4=0.75
Your question: When the number of happy people is equal to 1, can it be calculated by C (41) * C (12,2)/C (16,3)?
cannot
The premise of the first question is different from that of the second question. In the first question, c (1, 4) c (2, 12)/c (3, 16) is the probability that one person in 16 is a happy person.
The second question, the happiness probability of 1 person is randomly selected from the sample data of 16 people.