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.
Answer:
h=x-11/x
or
x=-11/h-1
Step-by-step explanation:
Answer:
Step-by-step explanation:
aha...
Answer:
32x10^4 and 3.2x10^5
Step-by-step explanation:
Mind marking me brainliest? :)
