Answer:
P(Y=1|X=3)=0.125
Step-by-step explanation:
Given :
p(1,1)=0
p(2,1)=0.1
p(3,1)=0.05
p(1,2)=0.05
p(2,2)=0.3
p(3,2)=0.1
p(1,3)=0.05
p(2,3)=0.1
p(3,3)=0.25
Now we are supposed to find the conditional mass function of Y given X=3 : P(Y=1|X=3)
P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)
P(X=3)=p(3,1) +p(3,2) +p(3,3)
P(X=3)=0.05+0.1+0.25=0.4

Hence P(Y=1|X=3)=0.125
Any number ends with 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9
Given any number. The square of this number is the last digit of square of the original numbers units digit.
For example
23*23 ends with 9 (3*3=9)
149*149 ends with 1 (9*9=81)
2564*2564 ends with 6 (4*4=16)
and so on
so all the possible unit digits of a square number are {0, 1, 4, 5, 6, 9}
because:
0*0= 0 ; 1*1=1; 2*2=4; 3*3=9; 4*4=16; 5*5=25, 6*6=36; 7*7=49; 8*8=64, 9*9=81
Thus, the probability that the square of a number selected from any set of numbers being 7, is 0.
Answer: 0
Answer:

Step-by-step explanation:

Answer:
I think that gives remind us of how pictures lokked back then and because that is what they used back then and it tells me is old
Step-by-step explanation: