In a certain factory, the production line which produces...

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BTGmoderatorLU: Moderator; Posts: 2505; Joined: Sun Oct 15, 2017 1:50 pm; Followed by:6 members

In a certain factory, the production line which produces...

by BTGmoderatorLU » Thu Jan 04, 2018 8:34 am

In a certain factory, the production line which produces the bags has the probability that a bag selected at random is defective is 0.02. If 6 bags are selected at random, what is the probability that at least one bag is defective?

$$A.\ 0.02^6$$
$$B.\ \left(0.02\right)\left(0.98\right)^6$$
$$C.\ \ 1-\left(0.02\right)^6$$
$$D.\ \ 0.98^6$$
$$E.\ \ 1-0.98^6$$

The OA is E.

I'm really confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

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Source: Beat The GMAT — Problem Solving | Thu Jan 04, 2018 8:34 am

GMAT/MBA Expert

Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Thu Jan 04, 2018 8:43 am

LUANDATO wrote:In a certain factory, the production line which produces the bags has the probability that a bag selected at random is defective is 0.02. If 6 bags are selected at random, what is the probability that at least one bag is defective?

$$A.\ 0.02^6$$
$$B.\ \left(0.02\right)\left(0.98\right)^6$$
$$C.\ \ 1-\left(0.02\right)^6$$
$$D.\ \ 0.98^6$$
$$E.\ \ 1-0.98^6$$

We want P(select at least 1 defective bag)
When it comes to probability questions involving "at least," it's best to try using the complement.
That is, P(Event A happening) = 1 - P(Event A not happening)
So, here we get: P(getting at least 1 defective bag) = 1 - P(not getting at least 1 defective bag)
What does it mean to not get at least 1 defective bag? It means getting zero defective bags.
So, we can write: P(getting at least 1 defective bag) = 1 - P(getting zero defective bags)

ASIDE: If P(bag is defective) = 0.02, then P(bag is NOT defective) = 0.98

P(getting zero defective bags)
P(getting zero defective bags) = P(1st bag is good AND 2nd bag is good AND 3rd bag is good AND 4th bag is good AND 5th bag is good AND 6th bag is good)
= P(1st bag is good) x P(2nd bag is good) x P(3rd bag is good) x P(4th bag is good) x P(5th bag is good) x P(6th bag is good)\
= 0.98 x 0.98 x 0.98 x 0.98 x 0.98 x 0.98
= 0.98^6

So, P(select at least 1 defective bag) = 1 - 0.98^6
= E

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GMAT/MBA Expert

Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Mon Sep 02, 2019 6:06 pm

BTGmoderatorLU wrote:In a certain factory, the production line which produces the bags has the probability that a bag selected at random is defective is 0.02. If 6 bags are selected at random, what is the probability that at least one bag is defective?

$$A.\ 0.02^6$$
$$B.\ \left(0.02\right)\left(0.98\right)^6$$
$$C.\ \ 1-\left(0.02\right)^6$$
$$D.\ \ 0.98^6$$
$$E.\ \ 1-0.98^6$$

The OA is E.

I'm really confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.

We are given that the probability that a selected bag is defective is 0.02, so we can determine that the probability of a non-defective bag is 1 - 0.02 = 0.98. We need to determine the probability that when 6 bags are selected at least one bag is defective.

We know the following:

1 = P(at least one defective bag is selected) + P(no defective bags are selected)

Thus,

P(at least one defective bag is selected) = 1 - P(no defective bags are selected)

Therefore, we need to determine P(no defective bags are among the 6 selected bags).

P(no defective bags are among the 6 selected bags) = 0.98^6

Thus, P(at least one defective bag is among the 6 selected bags) = 1 - 0.98^6

Answer: E

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