Question 1199785: A jar contains 32 red marbles numbered 1 to 32 and 50 blue marbles numbered 1 to 50. A marble is drawn at random from the jar. Find the probability of the given event. Please enter reduced fractions.
(a) The marble is red. P(red)=
(b) The marble is odd-numbered. P(odd)=
(c) The marble is red or odd-numbered. P(red or odd) =
(d) The marble is blue or even-numbered. P(blue or even) =
Answer by math_tutor2020(3816)   (Show Source):
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Part (a)
32 red + 50 blue = 82 total
P(red) = (number of red)/(number total)
P(red) = (32)/(82)
P(red) = (2*16)/(2*41)
P(red) = 16/41
Answer: 16/41
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Part (b)
R = set of red marbles labeled 1 to 32
B = set of blue marbles labeled 1 to 50
R = {1,2,3,...,31,32}
B = {1,2,3,...,49,50}
There are 32/2 = 16 odd values in set R.
There are 50/2 = 25 odd values in set B.
There are 16+25 = 41 odd values total.
41/82 = 1/2 of the values are odd.
Answer: 1/2
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Part (c)
The notation n(A) refers to "the number of items inside set A".
Example: A = {4,5,6} leads to n(A) = 3.
n(odd) = number of odd values
n(odd) = 41
This was calculated in part (b)
n(red) = number of red marbles
n(red) = 32
n(red and odd) = 16
This was calculated in part (b)
n(red or odd) = number of red or odd or both
n(red or odd) = n(odd) + n(red) - n(red and odd)
n(red or odd) = 41 + 32 - 16
n(red or odd) = 57
Phrased another way: There are 32 red + 25 odd blue = 57 marbles that are red or odd or both.
P(red or odd) = n(red or odd)/n(total)
P(red or odd) = 57/82
Answer: 57/82
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Part (d)
n(blue) = 50
n(even) = 41
n(blue and even) = 25
n(blue or even) = n(blue) + n(even) - n(blue and even)
n(blue or even) = 50 + 41 - 25
n(blue or even) = 66
Put another way: There are 50 blue + 16 red even = 66 marbles that are blue or even or both
P(blue or even) = n(blue or even)/n(total)
P(blue or even) = 66/82
P(blue or even) = 33/41
Answer: 33/41
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