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

Question

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}

CrackverbalGMAT · Answer

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

EgmatQuantExpert · Answer

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

GMATWhizTeam · Answer

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.

ScottTargetTestPrep · Answer

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

Basshead · Answer

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

bumpbot · Answer

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

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