Respuesta :
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:
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
=[tex]\frac{n[GL]}{n[T]}×\frac{n[GL]-1}{n[T]-1}[/tex]
=10/30*9/29=3/29
II.Both are good fruits
=[tex]\frac{n[BL]}{n[T]}×\frac{n[BL]-1}{n[T]-1}[/tex]
=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