hi
1st draw: green
2/12. (2 green out of 12 marbles), reduces to 1/6
2nd draw: blue
4/11 (4 blue marbles out of the 11 marbles remaining after 1st draw.)
Probability of theses BOTH happening: multiply these individual probabilities.
Conditional probability:
P(a and b) = P(a)P(b|a)
Prob(event 1) × Prob(event 2 given event 1).
1/6 • 4/11 = 4/66 = 2/33 or 0.0606. just a hair over 6% probability of the described sequence hapening.
First term(a)=2
D=8
Now
t13=a*12D
=2*12*8=192
Answer= 192
The answer is -x+37 .......
Answer: it is 4i78f4950
Step-by-step explanation:
