In a certain lottery, the probability that a number between 12 and 20,

Question

In a certain lottery, the probability that a number between 12 and 20, inclusive, is drawn is 1/6 . If the probability that a number 12 or larger is drawn is 2/3 , what is the probability that a number less than or equal to 20 is drawn?

(A) 1/18
(B) 1/6
(C) 1/3
(D) 1/2
(E) 5/6

(A) 1/18
(B) 1/6
(C) 1/3
(D) 1/2
(E) 5/6

Lucky2783 · Accepted Answer

Total numbers in lottery = 54 
1/3 rd are less than 12 . i.e 18 are less than 12. 
So total numbers less than or equal to 20 are 18 + 9 (12 to 20 including )

27/54 answer 1/2

KarishmaB · Answer

You can simply use sets concept in this question. The formula 
Total = n(A) + n(B) - n(A and B) is applicable here too.

Set 1: Number 12 or larger
Set 2: Number 20 or smaller

1 = P(Set 1) + P(Set 2) - P(Set 1 and Set 2) (combined probability is 1 because every number will be either "12 or more" OR "20 or less" OR both)
2/3 + P(Set 2) - 1/6 = 1
P(Set 2) = 1/2

Answer (D)

vishwaprakash · Answer

Total numbers between 12 and 20 including both = 9
and the probability that a number between 12 and 20, inclusive, is drawn is 1/6

Let X be the total number of lottery
\frac{9}{X} = \frac{1}{6}
Hence X = total number of lottery = 54.

the probability that a number 12 or larger is drawn = 2/3
Hence Number of lottery more than 12 = \frac{(2*54)}{3} = 36
Hence Number of Lottery below 12 = 54-36 = 18
Number of lotteries below 20 (including 20) = 18 + 9 = 27

hence Required probability = Number of lotteries below 20 (including 20) / the total number of lottery = 27/54 = 1/2

vishwaprakash · Answer

VeritasPrepKarishma Wow excellent explanation. thanks ... Will try to use your method if i encounter similar question anywhere

IanStewart · Answer

If 2/3 of the numbers are at least 12, and 1/6 of the numbers are between 12 and 20 inclusive, then 2/3 - 1/6 = 1/2 of all the numbers are strictly greater than 20. So 1/2 of the numbers must also be 20 or less, which is what the question is asking us to find.

luisdicampo · Answer

Deconstructing the Question  
Let:
-  A  = “the number is between 12 and 20 inclusive”, with  P(A)=1/6 
-  B  = “the number is 12 or larger”, with  P(B)=2/3

We are asked to find:
 P(	ext{number} \le 20)

Step-by-step  
Numbers that are  \ge 12  (event  B ) fall into two  disjoint  parts:
1) between 12 and 20 inclusive (event  A )
2) greater than 20 (event  >20 )

So we can write:
 B = A \cup (>20) 
and since they are disjoint:
 P(B)=P(A)+P(>20)

Solve for  P(>20) :
 P(>20)=P(B)-P(A)=\frac{2}{3}-\frac{1}{6}

Common denominator:
 \frac{2}{3}=\frac{4}{6} 
So:
 P(>20)=\frac{4}{6}-\frac{1}{6}=\frac{3}{6}=\frac{1}{2}

Now use the complement:
 P(	ext{number} \le 20)=1-P(>20)=1-\frac{1}{2}=\frac{1}{2}

Answer:     (D) 1/2

In a certain lottery, the probability that a number between 12 and 20,

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