Respuesta :
Answer: 6/11 and 10/21
Step-by-step explanation:
A bag of marbles has 10 green marbles and 12 black marbles. The probability of picking a black marble on the first try will be:
Green marbles= 10
Black marbles= 12
Total marbles= 12+10 = 22
Probability of picking a black marble = 12/22 = 6/11
If the marble is not replace, the probability of picking a green marble on the second try will be:
Green marbles= 10
Black marbles= 12-1= 11
Total marbles= 10+11= 21
Probability of picking a green marble= 10/21
Answer:
20/77 or 0.26 or 26%
Step-by-step explanation:
Extracting the key information from the question:-
*** There's a bag containing 10 green and 12 black marbles.
*** We are simply required to find the probability of picking a black marble first and then a green marble(NOT THE RESPECTIVE CHANCES).
The chance of picking a black marble is:
Number of black marbles/total number of marbles
= [tex]\frac{12}{12+10}[/tex]= 12/22
The chance of picking a green marble is:
Number of green marbles/total number of marbles
= [tex]\frac{10}{10+12}[/tex]
= 10/22
Note that when a marble is picked without replacing it, the total number of marbles available for further picking will reduce by one.
Now, without replacement, the probability of picking a black marble on the first attempt and a green marble on the second attempt is:
P(black, then green)[without replacement]
= [tex]\frac{12}{22} *\frac{10}{21}[/tex]
= 120/462
= 20/77 or 0.26 or 26%
