Question

1.If a number is chosen at random from the integers 5 to 25 inclusive , find the probability that the number is a multiple of 5 or 3.

2.Good Limes =10
Good Apples = 8
Bad Limes = 6
Bad Apples 6
The information above shows the number of limes and apples of the same size in a bag . If two of the fruits are picked at random , one at a time without replacement .Find the probability that :
I. Both are good limes
II.Both are good fruits
III. One is a good apple and the other a bad lime

Answer 1

I'll do problem 1 to get you started.

set A = multiples of 3 between 5 and 25 = {6, 9, 12, 15, 18, 21, 24}

there are 7 items in set A, so we can say n(A) = 7

set B = multiples of 5 between 5 and 25 = {5,10,15,20,25}

Here we have n(B) = 5

set C = multiples of 3 and 5, between 5 and 25 = {15}

n(C) = 1 which we can rewrite as n(A and B) = 1.

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To summarize so far,

• n(A) = 7
• n(B) = 5
• n(A and B) = 1

From those three facts, then we can say,

n(A or B) = n(A) + n(B) - n(A and B)

n(A or B) = 7 + 5 - 1

n(A or B) = 11

There are 11 values between 5 and 25 that are multiples of 5, multiples of 3, or both.

Those 11 values are: {5, 6, 9, 10, 12, 15, 18, 20, 21, 24, 25}

This is out of 25-5+1 = 21 values overall which are in the set {5,6,7,...,24,25}

So we have 11 values we want out of 21 overall, which leads to the probability 11/21

Final Answer: 11/21

Answer 2

Solution given:

total outcomes between 5 to25 inclusive

n[T]=25-5+1=21

multiple of 5n[5]=5,10,15,20,25=5

multiple of 3n[3]=6,9,12,15,18,21,24=7

now

probability of getting multiple of 5p[5]=5/21

and

probability of getting multiple of 3 p[3]=7/21=1/3

again

the probability that the number is a multiple of 5 or 3 P[5or 3]=p[5]+p[3]=5/21+1/3=4/7

the probability that the number is a multiple of 5 or 3 is 4/7.

2:

.Good Limes n[GL] =10

Good Apples n[GA]= 8

Bad Limes n[BL] = 6

Bad Apples n[BA]= 6

total fruits n[T]=10+8+6+6=30

no of good apple n[G]=10+8=18

no of bad apple n[B]=6+6=12

again

I. Both are good limes

=
`(n[GL])/(n[T])×(n[GL]-1)/(n[T]-1)`

=10/30*9/29=3/29

II.Both are good fruits

=
`(n[BL])/(n[T])×(n[BL]-1)/(n[T]-1)`

=6/30*5/29=1/29

III. One is a good apple and the other a bad lime

=n[G]/n[T] *n[B]/(n[T]-1)

=18/30*12/29=36/145