There are 10 bottles wine in a carton. Three bottles will be selected

Question

There are 10 bottles wine in a carton. Three bottles will be selected and examined at random. If any of the three bottles is defective, the carton will be sent
23 back. If a carton contains exactly three defective bottles, what is the probability that the carton will not be sent back?

(A) 1/9
(B) 3/10
(C) 7/10
(D) 7/24
(E) 17/24

(A) 1/9
(B) 3/10
(C) 7/10
(D) 7/24
(E) 17/24

MBA20 · Answer

What is wrong in my solution -

1 - 3C1/10C3

LevanKhukhunashvili · Answer

We need to pick 3 bottles
None of them should be defective

The probability to pick first non-defective bottle -->  \frac{7}{10} 
The probability to pick second non-defective bottle -->  \frac{6}{9}  (as we have already picked one)
The probability to pick third non-defective bottle -->  \frac{5}{8}

We want to appear first AND second AND third cases, so we have to multiply probabilities  \frac{7}{10} * \frac{6}{9} * \frac{5}{8} = \frac{7}{24}

IMO
Ans:  D

anmolgmat14 · Answer

Why 1-3c1/10c3 is wrong ?

MBA20 · Answer

VeritasKarishma  Can u Help ?

aab01 · Answer

The answer can also be solved using combinations.

Number of ways the bottles can be selected = 10C3 => 120
Number of ways only the non-defective bottles are selected = 7C3 => 35 (since there are 3 defective bottles in this carton)
Probability of carton not being sent back = 7C3 / 10C3 => 35/120 => 7/24

KarishmaB · Answer

If you are calculating the opposite probability, you need to calculate the number of ways of selecting "1 defective bottle and 2 non defective ones" or 2 defective bottles and 1 non defective one" or "3 defective bottles"

This can be calculated as:

/ 10C3

/ (10*9*8/6)

85*6/10*9*8

17/24

Then, no defective bottle = 1 - 17/24 = 7/24

Mugdho · Answer

KarishmaB

If we want to do it manually like- 
1 defective bottle and 2 non defective ones
(3/10)*(7/9)*(6/8)*3!

Can we write it? As every bottle is distinct and we could take 2 non-defective bottle first and then the defective one...shouldn’t we multiply it by 3!?

Posted from my mobile device

KarishmaB · Answer

We multiple it by 3, not 3!

There are two different types of bottles - 3 defective, 7 non defective. To us in the question, the only difference is that 3 are defective and 7 non defective. 
Think about the number of ways in which you could get exactly one defective bottle. 
You could pick DNN or NDN, or NND. 
That is 3 ways.

dave13 · Answer

KarishmaB

1) the probability of choosing at least 1 defective bottle = (3*7*6)/(10*9*8)= 7/40; * 3!/2! = 21/40
2) the probability of choosing at least 2 defective bottles = (3*2*7)/(10*9*8) = 7/120 * 3!/2! = 21/120
3) the probability of choosing at least 3 defective bottles = (3*2*1)/(10*9*8) = 1/120 * 3!/3! = = 1/120

1 - (21/40+21/120+1/120 ) = 17/24

What`s wrong with my solution ?

KarishmaB · Answer

Your calculations are correct except that 
(21/40+21/120+1/120 ) = 17/24 and 
1 - (21/40+21/120+1/120 ) = 7/24

But note that you are calculating the probabilities of choosing "exactly one defective bottle" and "exactly two defective bottles" and "exactly three defective bottles," not at least. 'the probability of choosing at least 1 defective bottle' would include the probabilities of choosing 2 or 3 defective bottles.

There are 10 bottles wine in a carton. Three bottles will be selected

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There are 10 bottles wine in a carton. Three bottles will be selected

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