A jar contains 4 red marbles numbered 1 to 4 and 7 blue marbles numbered 1 to 7.
ID: 3199258 • Letter: A
Question
A jar contains 4 red marbles numbered 1 to 4 and 7 blue marbles numbered 1 to 7. A marble is drawn at random from the jar a) Find the probability that the marble is red or odd-numbered. Enter your answer as a percent, rounded to the nearest tenth of a percent. 72.7 % bj Find the probability that the marble is blue, given that the marble is even-numbered. Enter your answer as a percent, rounded to the nearest tenth of a percent 60 c) Find the probability that the marble is blue and odd-numbered Enter your answer as a percent, rounded to the nearest tenth of a percent. 94%Explanation / Answer
(a). Total number of marbles = 4 red + 7 blue = 11
Probability of red marble = number of red marbles / total number of marbles = 4/11
Even number marbles :
Red : 2,4
Blue : 2,4,6
So total 5 marbles
Odd number marbles :
Red : 1,3
Blue : 1,3,5,7
Total 6 odd number marbles
Probability of odd number marble = 6/11
P(A or B) = P(A)+P(B)
= 4/11 + 6/11 = 10/11
= 0.909 = 90.9%
(b)
Probability of blue marble = 7/11
Probability of even in blue marble = 3/7
P(A and B) = P(A) * P(B)
= (7/11)*(3/7) = 21/77
= 0.2727 = 27.3%
(c)
Probability of blue = 7/11
Probability of odd in blue = 4/7
Probability of blue and odd = (7/11)*(4/7)
= 28/77 = 0.3636
. = 36.4%
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