A box contains 100 tickets marked 1 to 100. One ticket is

Question

A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?

A. 83/100
B. 67/100
C. 51/100
D. 49/100
E. 33/100

The official answer is:

51/100

Bunuel · Accepted Answer

A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?

Total = 100 numbers.

Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C

Now,  both  A (numbers divisible by 2) and B (numbers divisible by 3) contain C (numbers divisible by 6) so to get the numbers which are divisible by  either 2 or 3, but not 6  we should subtract C twice, once for A and once for B: A + B - 2*C= 51

Probability = 51/100

If it were: "What is the probability that the number on the ticket will be divisible by either 2 or 3 (or both)?" Then we would subtract C only once to get rid of double counting.

fluke · Answer

I thought it to be 67/100 and realized my mistake.

Count of numbers divisible by 2 = 50. In these 50 numbers there are 16 numbers that are divisible by 6 and we must take them out. 50-16=34

Count of numbers divisible by 3 = 33. In these 33 numbers there are 16 numbers that are divisible by 6 and we must take them out. 33-16=17

Total numbers divisible by 2 or 3 but not 6 = 34+17=51

Probability = 51/100.

Good one.

nonameee · Answer

Bunuel, thanks. I got it.

MHIKER · Answer

thanks to bunuel and fluke

jlgdr · Answer

Every 6 numbers there are 3 divisible by 2 or 3 and not divisible by 6 Then in 100 numbers there are 16 sets of 6 and 4 numbers left as remainder. So, 16(3) = 48. Now of the numbers remaining (92,98,99 and 100) there 3 of them are multiples of 2 or 3 but not 6. So total is 51 numbers Hence prob is 51/100 Hope it helps Cheers J

Enael · Answer

Is it the same as saying: divisible by either 2 or 3 but not by both?

Bunuel · Answer

_______
Yes, it is the same.

Alphatango · Answer

51/100 wud be the soln

Divyadisha · Answer

Total numbers= 100
numbers divisible by 2= 50
numbers divisible by three= 33
numbers divisible by 6= 16

numbers divisible by 2 and 3 but not 6= 50-16+33-16= 51

probability= 51/100

utkarshthapak · Answer

Please update the options for this question. 
Correct Ans should be 51/100.
No. of multiples of 2 = 50
No. of multiples of 3 = 33
No. of multiples of 6 = 16
Since both multiples of 2 & 3 will also contain multiples of 6 so we need to subtract it twice.
Total no. of multiples of 2 & 3 but not 6 = 51.
Probability = 51/100

cavana · Answer

1) 100/2=50
2) 100/3=33
3) 100/3*2=16
4) 50-16+33-16=51

ShivamoggaGaganhs · Answer

We can use a venn diagram for this

Divisible by both 2 and 3 means

Divisible by 6

AJ0784 · Answer

Numbers divisible by 2
=100/2= 50
Numbers divisible by 3
=100/3= 33
Numbers divisible by 6
= 100/6=16
Now,Numbers divisible by 2 but not by 6= 50-16=34

And,Numbers divisible by 3 but not by 6= 33-16=17

Therefore,Numbers divisible by 2 and 3 but not by 6= 34+17=51

Probability = 51/100=0.51

Posted from my mobile device

bv8562 · Answer

Total numbers= 100
Numbers divisible by 2 = 50
Numbers divisible by 3 = 33
Numbers divisible by 6 = 16

(Refer to the attached screenshot for the venn diagram)

Numbers divisible by either 2 or 3 but not 6 = 34+17 = 51

Probability = Favorable options/Total options

Favorable options = 51C1 = 51
Total options = 100C1 = 100

P(Number divisible by either 2 or 3 but not 6)  =  51/100

bumpbot · Answer

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A box contains 100 tickets marked 1 to 100. One ticket is

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