A bag contains 10 red, 5 yellow, and 6 blue marbles. What is the probability of pulling out one blue marble and then one red marble without replacing the first blue marble?
Answers
Answered by
2
Answer:
0.2857;0.67
Step-by-step explanation:
total no. of marbles in bag = n(s)=10+5+6 = 21
no. of blue marbles = n(e) = 6
P(blue marbles)= n(e)/n(s)
P(E)=6/21
=0.2856
Now,
no. of marbles left in bag = 21-6 = 15
no. of red marbles = 10
P(red marbles) = 10/15
P(E)=0.67
Answered by
0
Answer:
Step-by-step explanation:
Total marbles= 21
Red marbles = 10
Yellow marbles = 5
Blue marbles= 6
Probability of pulling one blue marble = 6/21 = 2/7
Probability of pulling one red marble without replacing the first blue marble = 10/20 =1/2
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