In a certain game, a player begins with a bag containing tiles numbere

Question

In a certain game, a player begins with a bag containing tiles numbered 1 through 10, each of which has an equal probability of being selected. The player draws one tile. If the tile is even, the player stops. If not, the player draws another tile without replacing the first. If this second tile is even, the player stops. If not, the player draws a third tile—without replacing either of the first two tiles—and then stops. What is the probability that at the conclusion of the game, the sum of the tiles that the player has drawn is odd?

A) 5/18
B) 13/36
C) 3/8
D) 5/8
E) 23/36

Kudos for a correct solution.

A) 5/18
B) 13/36
C) 3/8
D) 5/8
E) 23/36

Kudos for a correct  solution .

Psiva00734 · Accepted Answer

hi bunel pls correct me if my approach is wrong.
player begins with containing tiles 1 to 10 and have equal probability of selecting means 
Proabability of selecting one number =1/10. here 5 even numbers and 5 odd numbers are there.

Next , player draws one title , if number is even player stops or otherwise title is odd without replacement
player draws second title.
If second title is even , player stops or title is odd without replacement player draws third title.

in third title , without replacement of first and second title, player draws and stops it.

the sum of tilte probability is odd. here two conditions are possible.
1st condition is
1st title is odd+ 2nd title is even stops= probability of selecting one title is 1/10*5c1.
Here are we are not selecting 1st condition as even stops because sum of tile is odd.
Here 5 odd numbers are there we can select 1out of 5 odd numbers.
 without replacement of fist we select second tilte is even. is 5/10*5c1/9c1. 
here we are selecting one number out of remaining 9 numbers. so probability is 5/18.
we are selecting 1 even number out of 5.

2nd condition is 1stodd without replacement,2nd odd without replacement and 3rd one also odd to get odd as sum of title.
Then probability is 5/10*4/9*3/8=1/12.

Finally sum of probability of two conditions is 5/18+1/12
=13/36.
so option B is correct.

tvrs09 · Answer

Hi

Option B

There are two possible cases.

First case : Odd Odd Odd

Probability is (5/10*4/9*3/8) => 1/12

Second case : Odd even

Probability is (5/10*4/9) => 5/18

Total probability => 1/12 + 1/18 = 13/36

Am I missing anything?

Cheers!!

mejia401 · Answer

Odd, Odd, Odd: 5/10*4/9*3/8 = 1/12
Odd, Even: 5/10*4/9 = 65/180

1/12+65/180=13/36
Option B

The tricky part is picking up that only three tiles are drawn, very tricky wording, which I suppose is a classic technique for 700-type questions. Didn't see it myself at first.

Kr,
Mejia

suhasreddy · Answer

out of the four ways in which the game can end (E,OE,OOE,OOO) the two ways that result in an odd sum are OE,OOO.
so,the probability is P(OE)+P(OOO)= (5/10)* (5/9)+(5/10)*(4/9)*(3/8)=13/36.

PS:
P( OE ) = 5odd/total 10 * 5 even / remaining 9.
P( OOO ) = 5odd/total 10 * remaining 4odd/ total remaining 9 * remaining 3 odd / total remaining 8 .

Option B

Bowtye · Answer

I solved by first finding the probability that it would be even.

There are two options that the out come could be even.

Option 1:

First tile picked is even. Probability of 1/2

Option 2:

First two tiles picked are odd and the third tile is even. Probability of (1/2)*(4/9)*(5/8) = 5/36

1-(5/36+1/2) = 13/36

Option B

SwaroopKumar · Answer

Hi bunuel,

It should be 13/36 as explained by others.

But, if the tiles are replaced every time............

Is the answer for that question: (5C1 * 5C1/10C2)+(5C3/10C3)

First Case - First tile odd & Second tile even; Second Case - All the tiles odd.

Please confirm

Thanks,
SK

shriramvelamuri · Answer

Hi All,

Please enlighten me.

I see 3 scenarios.

1. 3 cards are all odd. Hence.

5/10 * 4/9 * 3/8 = 1/12;

2. One odd and one even; (The game ended when the even card was picked)

5/10 * 5/ 9 = 5 /18;

3. Two Odds and one even (the game finished)

5/10 * 4 /9 * 5/8= 5/36.

hence probability is coming 1/2.

Please help me with this question.

ersheet · Answer

Hi shriramvelamuri,

the question is asking for "the probability that at the conclusion of the game, the sum of the tiles that the player has drawn is odd"
OOE will make the sum even and hence that condition is not considered!!!

Bambaruush · Answer

My solution has totally different result. I beg at least a little clue from Bunuel

. There are following number of ways: 1. E 2. O+E=O 3. O+O+E=E 4. O+O+O+E=O 5. O+O+O+O+E=E 6. O+O+O+O+O+E=O We need only odd results. 2. 5/10 * 5/9 = 5/18 4. 5/10 * 4/9 * 3/8 * 5/7 = 5/84 6. 5/10 * 4/9 * 3/8 * 2/7 * 1/6 * 5/5 = 1/252 Total probability is 5/18 + 5/84 + 1/252 = 67/126

mathivanan · Answer

the game concludes in one of the following situations:
a)first tile is even-number of tiles drawn is 1
b)first odd and second even- number of tiles drawn is 2
c) first odd, second odd, and third even- 3 tiles
d) first odd, second odd, third odd and fourth even-4 tiles
e)first odd, second odd, third odd, fourth odd, and fifth even-5 tiles
f)first odd, second odd, third odd, fourth odd, fifth odd, and sixth even-6 tiles
No more draw is required.
Is it sum of the tiles or sum of the numbers on the tiles. In both the cases, the probability is 1/2

Bambaruush · Answer

OH MY GOD. game ends after 3 draws. hitting my head.

Careless not to see above comments.

alexanderhle · Answer

Two distinct events:

1. odd, odd, odd: P = 5/10 * 4/9 * 3/8 = 1/12
2. odd, even: P = 5/10 * 5/9 = 5/18; 5/9 implies that you still have 5 even tiles left from the 9 total tiles (remember that 1 odd tile was picked up in the first round)

Summing the two distinct events: 1/12 + 5/18 = 13/36

Answer: B.

bumpbot · Answer

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In a certain game, a player begins with a bag containing tiles numbere

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