NEW HERE? LEARN MORE
closeclose
Last visit was: 30 Apr 2024, 03:20 It is currently 30 Apr 2024, 03:20
Decision Tracker
My Rewards
New posts
New comers' posts

Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

A bag contains three flavors of gum: p pieces of peppermint gum, s pie

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
avatar
S
sarphant123
Manager
Manager
Joined: 13 Oct 2019
Posts: 67
Own Kudos [?]: 218 [1]
Given Kudos: 191
Send PM
A bag contains three flavors of gum: p pieces of peppermint gum, s pie [#permalink] New post   26 Jul 2020, 23:32
1
Bookmarks
00:00
A
B
C
D
E

Difficulty:

25% (medium)

Question Stats:

79% (01:51) correct 21% (01:56) wrong based on 178 sessions
Hide Show timer Statistics
A bag contains three flavors of gum: p pieces of peppermint gum, s pieces of spearmint gum, and w pieces of wintergreen gum. Two peppermint and three wintergreen pieces of gum are removed from the bag and are not replaced. If Jane then randomly chooses one piece of gum from the bag, what is the probability that the piece of gum Jane picks is not spearmint?
(A)\(\frac{s}{(p-2)+(w-3)}\)
(B)\(\frac{p+w-5}{p+w+s}\)
(C)\(\frac{1-s}{p+w+s-5}\)
(D)\(\frac{1}{s}\)
(E)\(\frac{p+w-5}{p+w+s-5}\)
ShowHide Answer
Official Answer
E
User avatar
L
CrackverbalGMAT
GMAT Club Legend
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4946
Own Kudos [?]: 7631 [1]
Given Kudos: 215
Location: India
Send PM
Re: A bag contains three flavors of gum: p pieces of peppermint gum, s pie [#permalink] New post   27 Jul 2020, 04:04
1
Bookmarks
Top Contributor
Probability = Favorable Outcomes / Total Outcomes

Also nC1 = n, nC0 = nCn = 1

When 2 peppermints and 3 wintergreens are removed, the total pieces left in the bag = (p - 2) + (w - 3) + s = p + w + s - 5

Total Outcomes = \(p + w + s - 5^\)C\(_1\) = p + w + s - 5

When she picks 1 piece, it should not be spearmint, i.e she has to pick from peppermint or wintergreens.

Number of peppermints and wintergreens after removal = (p - 2) + (w - 3) = p + w - 5

Favorable Outcomes =\({p + w - 5}^\)C\(_1\) = p + w - 5

The required probability = \(\frac{p + w - 5 }{ p + w + s - 5}\)

Option E

Arun Kumar
User avatar
L
EgmatQuantExpert
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3723
Own Kudos [?]: 16884 [0]
Given Kudos: 165
Send PM
Re: A bag contains three flavors of gum: p pieces of peppermint gum, s pie [#permalink] New post   30 Jul 2020, 07:46
Expert Reply
Given

    • A bag contains three flavors of gum: p pieces of peppermint gum, s pieces of spearmint gum, and w pieces of wintergreen gum.
    • Two peppermint and three wintergreen pieces of gum are removed from the bag and are not replaced.
    • Jane then randomly chooses one piece of gum from the bag.

To Find

    • The probability that the piece of gum Jane picks is not spearmint.

Approach and Working Out

    • Total number of gums = p + s + w
    • After removing 2 peppermint and three wintergreen number of pieces = p + w + s – 5.
      o Number of spearmints remains s.
      o Number of peppermint and wintergreen is p + w – 5.
    • Required probability = \(\frac{(p + w – 5)}{(p + w + s -5)}\)

Correct Answer: Option E
User avatar
L
GMATWhizTeam
GMATWhiz Representative
Joined: 07 May 2019
Posts: 3409
Own Kudos [?]: 1802 [0]
Given Kudos: 68
Location: India
GMAT 1: 740 Q50 V41
GMAT 2: 760 Q51 V40
Send PM
Re: A bag contains three flavors of gum: p pieces of peppermint gum, s pie [#permalink] New post   30 Jul 2020, 14:24
Expert Reply
sarphant123 wrote:
A bag contains three flavors of gum: p pieces of peppermint gum, s pieces of spearmint gum, and w pieces of wintergreen gum. Two peppermint and three wintergreen pieces of gum are removed from the bag and are not replaced. If Jane then randomly chooses one piece of gum from the bag, what is the probability that the piece of gum Jane picks is not spearmint?
(A)\(\frac{s}{(p-2)+(w-3)}\)
(B)\(\frac{p+w-5}{p+w+s}\)
(C)\(\frac{1-s}{p+w+s-5}\)
(D)\(\frac{1}{s}\)
(E)\(\frac{p+w-5}{p+w+s-5}\)


Solution


    • Number of pieces of
      o peppermint = p
      o spearmint = s
      o winter green = w
    • Now, 2 pieces of peppermint gum and 3 pieces of wintergreen gum are removed from the bags.
      o So new number of pieces of
         Peppermint = p-2
         Spearmint = s
         Winter green = w – 3
    • Thus, total number of pieces of gums in the bag \(= (p – 2) + s + (w – 3) = p + w + s - 5\)

    • Number of pieces of gums which are not spearmint \(= (p – 2) + (w – 3)= p + w – 5\)

    • Therefore, the probability of picking a non-spearmint gum \(= \frac{p + w -5}{p +w +s -5}\)
Thus, the correct answer is Option E.
User avatar
L
ScottTargetTestPrep
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 18775
Own Kudos [?]: 22092 [0]
Given Kudos: 283
Location: United States (CA)
Send PM
Re: A bag contains three flavors of gum: p pieces of peppermint gum, s pie [#permalink] New post   02 Aug 2020, 09:44
Expert Reply
sarphant123 wrote:
A bag contains three flavors of gum: p pieces of peppermint gum, s pieces of spearmint gum, and w pieces of wintergreen gum. Two peppermint and three wintergreen pieces of gum are removed from the bag and are not replaced. If Jane then randomly chooses one piece of gum from the bag, what is the probability that the piece of gum Jane picks is not spearmint?
(A)\(\frac{s}{(p-2)+(w-3)}\)
(B)\(\frac{p+w-5}{p+w+s}\)
(C)\(\frac{1-s}{p+w+s-5}\)
(D)\(\frac{1}{s}\)
(E)\(\frac{p+w-5}{p+w+s-5}\)


Solution:

When Jane chooses one piece of gum randomly, there are a total of p - 2 + w - 3 + s = p + w + s - 5 pieces of gum in the bag and p - 2 + w - 3 = p + w - 5 pieces of gum that are not spearmint. Therefore, the probability is (p + w - 5) / (p + w + s - 5).

Answer: E
User avatar
G
Basshead
Director
Director
Joined: 09 Jan 2020
Posts: 964
Own Kudos [?]: 223 [0]
Given Kudos: 434
Location: United States
Send PM
Re: A bag contains three flavors of gum: p pieces of peppermint gum, s pie [#permalink] New post   13 Oct 2020, 03:44
Total pieces of gum: p + s + w
5 pieces of gum are removed: p + s + w - 5
Not spearmint: p + w - 5

p + w - 5 / p + s + w -5
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32760
Own Kudos [?]: 823 [0]
Given Kudos: 0
Send PM
Re: A bag contains three flavors of gum: p pieces of peppermint gum, s pie [#permalink] New post   11 Nov 2021, 06:57
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
gmatclubot
Re: A bag contains three flavors of gum: p pieces of peppermint gum, s pie [#permalink]
11 Nov 2021, 06:57
Moderators:
Bunuel
Math Expert
93033 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions