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There are 3 boxes labeled A, B and C. Box A contains 6 coins, 1 of whi
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28 Aug 2022, 08:36
28 Aug 2022, 08:36
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Question Stats:
80% (02:01) correct
20% (02:05) wrong based on 5 sessions
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There are 3 boxes labeled A, B and C. Box A contains 6 coins, 1 of which is made of gold. Box B contains 4 coins, 1 of which is made of gold. Box C contains 2 coins, 1 of which is made of gold. If a box is randomly selected, and a coin is randomly selected from that box, what is the probability the selected coin is made of gold?
A) 1/12
B) 1/4
C) 11/36
D) 1/3
E) 5/12
A) 1/12
B) 1/4
C) 11/36
D) 1/3
E) 5/12
There are 3 boxes labeled A, B and C. Box A contains 6 coins, 1 of whi
[#permalink]
28 Aug 2022, 09:36
28 Aug 2022, 09:36
1
Case 1:-
short cut
1/3(1/6+1/4+1/2) = 11/36
Case 2:-
BoxA-6 coins (1G & 5NG)
BoxB-4 coins(1G & 3NG)
Boxc-2 coins (1G & 1NG)
P(not gold in box A) - 5/6
P(Not Gold in box B) - 3/4
P(not Gold in box C) - 1/2
All the cases
P(Not gold in all 3 boxes) = 5/6+3/4+1/2= 25/12
Randomly one box is chosen = 1/3
Now Probability of not getting gold from random boxes = (25/12)*1/3 = 25/36
Now, P(of Getting Gold Coin from random boxes) = 1-25/36 = 11/36
Answer is 11/36
short cut
1/3(1/6+1/4+1/2) = 11/36
Case 2:-
BoxA-6 coins (1G & 5NG)
BoxB-4 coins(1G & 3NG)
Boxc-2 coins (1G & 1NG)
P(not gold in box A) - 5/6
P(Not Gold in box B) - 3/4
P(not Gold in box C) - 1/2
All the cases
P(Not gold in all 3 boxes) = 5/6+3/4+1/2= 25/12
Randomly one box is chosen = 1/3
Now Probability of not getting gold from random boxes = (25/12)*1/3 = 25/36
Now, P(of Getting Gold Coin from random boxes) = 1-25/36 = 11/36
Answer is 11/36
Re: There are 3 boxes labeled A, B and C. Box A contains 6 coins, 1 of whi
[#permalink]
28 Aug 2022, 13:50
28 Aug 2022, 13:50
1
Independently-
Box 1- 1/6 chance of gold
Box 2- 1/4 chance of gold
Box 3- 1/2 chance of gold
Each box has a 1/3 chance of being chosen, so you have to multiply each chance of getting the gold by 1/3
Box 1 gold- 1/18 chance
Box 2 gold- 1/12 chance
Box 3 gold- 1/6 chance
Add them together with a common denominator, which is 36, and the total chance of getting gold ends up
2/36 + 3/36 + 6/36 = 11/36, C
Box 1- 1/6 chance of gold
Box 2- 1/4 chance of gold
Box 3- 1/2 chance of gold
Each box has a 1/3 chance of being chosen, so you have to multiply each chance of getting the gold by 1/3
Box 1 gold- 1/18 chance
Box 2 gold- 1/12 chance
Box 3 gold- 1/6 chance
Add them together with a common denominator, which is 36, and the total chance of getting gold ends up
2/36 + 3/36 + 6/36 = 11/36, C
