A box contains orange, green and blue balls. If one ball is chosen at

Solution

Step 1: Analyse Question Stem

• Let us assume that b, g, and o are the number of blue balls, green balls and orange balls in the box, respectively.
• If one ball is chosen at random from the box, we need to find the probability that the chosen ball is orange.

o Or, we need to find \(= \frac{o}{o+g+b}\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

Statement 1: The probability that the chosen ball is blue is one-fourth of the probability that the chosen ball is not blue

• Probability of not choosing a blue ball is same as probability of choosing either a green ball or an orange ball.
• So, according to this statement:\( 4*\frac{b}{o+g+b} = \frac{g+o }{ o+g+b }……………Eq.(i)\)
• We know that, \(\frac{b}{o+g+b} + \frac{g+o}{o+g+b}= 1 \)

o Thus, \(\frac{b}{o+g+b} + 4* \frac{b}{o+g+b }= 1\)
o Or, \(\frac{b}{o+g+b} = \frac{1}{5}\)
• Now, from Eq.(i) and the above equation, we have,

o \(\frac{g+o}{o+g+b} = \frac{4}{5} ⟹ \frac{g}{o+g+b }+ \frac{o}{o+g+b} = \frac{4}{5}\)
• However, we don’t know the probability of picking one green ball so we cannot find the required probability.

Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.

Statement 2: If there were 15 fewer orange balls in the box, the probability that the chosen ball is orange would have been equal to the probability that the chosen ball is blue

• According to this statement: \(\frac{o-15}{o-15 + g+b }= \frac{b}{o-15 + g+b}\)

o \(o -15 = b ⟹ o = b+15\)
• So, \(\frac{o}{o+g+b} = \frac{b}{o+g+b }+ \frac{15}{o+g+b}\)
• However, we neither know the value of b nor the value of o +g + b. So we cannot find the required probability.

Hence, statement 2 is also NOT sufficient and we can eliminate answer Option B.

Step 3: Analyse Statements by combining.

From statement 1:\(\frac{b}{o+g+b} = \frac{1}{5}\)
From statement 2: \(\frac{o}{o+g+b} = \frac{b}{b+g+b} + \frac{15}{o+g+b}\)
On combining both we have, \(\frac{o}{o+g+b} = \frac{1}{5} + \frac{15}{o+g+b}\)
• However, we still doesn’t know the value of o +g+b. So we cannot find the required probability.
Thus, the correct answer is Option E.

Alternate solution

Step 1: Analyse Question Stem

• Let us assume that b, g, and o are the number of blue balls, green balls and orange balls in the box, respectively.
• If one ball is chosen at random from the box, we need to find the probability that the chosen ball is orange.

o Or, we need to find \(= \frac{o}{o+g+b}\)

Step 2: Analyse Statements Independently (And eliminate options) – AD/BCE

Statement 1: The probability that the chosen ball is blue is one-fourth of the probability that the chosen ball is not blue

• Probability of not choosing a blue ball is same as probability of choosing either a green ball or an orange ball.
• So, according to this statement: \(4*\frac{b}{o+g+b} = \frac{g+o }{ o+g+b }……………Eq.(i)\)

o \(4b = o+g ⟹ o+g+b = 5b\)
• Therefore, \(\frac{o}{o+g+b} = \frac{o}{5b} \)

o However, we neither know the relation between o and b not the value of o and b. So we cannot find the required probability.

Hence, statement 1 is NOT sufficient and we can eliminate answer Options A and D.

Statement 2: If there were 15 fewer orange balls in the box, the probability that the chosen ball is orange would have been equal to the probability that the chosen ball is blue

• According to this statement: \(\frac{o-15}{o-15 + g+b} = \frac{b}{o-15 + g+b}\)

o \(o -15 = b ⟹ o = b+15\)
• So, \(\frac{o}{o+g+b} = \frac{b}{o+g+b} + \frac{15}{o+g+b}\)
• However, we neither know the value of b nor the value of o +g + b. So we cannot find the required probability.

Hence, statement 2 is also NOT sufficient and we can eliminate answer Option B.

Step 3: Analyse Statements by combining.

From statement 1: \(\frac{o}{o+g+b }= \frac{o}{5b}\)
From statement 2: \(o -15= b\)
On combining both we have, \(\frac{o}{o+g+b} = \frac{o}{5*(o-15)}\)

• However, we still doesn’t know the value of o. So we cannot find the required probability.

Thus, the correct answer is Option E.

We're picking only one ball, so the question is just asking "what fraction of the balls are orange?" The word "probability" is just a distraction.

Statement 1 tells us the ratio of blue to not-blue balls is 1 to 4, so 1/5 of the balls are blue.

Statement 2 just tells us the number of orange balls is 15 greater than the number of blue ones.

Without any other numbers, that is clearly insufficient; with tons of balls in the bag, then just slightly more than 1/5 will be orange (since the difference of 15 balls then won't matter much) but with very few balls in the bag, the fraction that are orange will be much larger than 1/5 (since then 15 extra orange balls will make a meaningful difference). But we can confirm: we might have 10 blue, 25 orange, and 15 green, and a 1/2 chance to pick an orange ball, or we might have 200 blue, 215 orange, and the 585 green and a 23/100 chance to pick an orange ball, among other possibilities. So the answer is E.

That said, I find the meaning of Statement 2 a bit ambiguous. I've taken it to mean "if you remove 15 orange balls, the probability of picking blue equals the probability of picking orange". I think because of the verb tenses used, one could also take it to mean "if you remove 15 orange balls, the probability of picking orange would be equal to the probability, before removing the 15 balls, of picking blue", which makes generating examples a bit more complicated, though doesn't change the answer. I suspect the interpretation I've made is the intended one in any case.

Orange, Green and Blue balls
For probability of picking Orange, we need to find the number of orange balls and the total number of balls.

(1) The probability that the chosen ball is blue is one-fourth of the probability that the chosen ball is not blue

So Blue: (Green + Orange) = 1 : 4
If we have B blue balls, we have 4B (Green + Orange) balls. Then we have 5B total balls. No info about orange balls. Not sufficient.

(2) If there were 15 fewer orange balls in the box, the probability that the chosen ball is orange would have been equal to the probability that the chosen ball is blue

This tells us that if there were 15 fewer Orange balls, the number of blue balls and number of orange balls would become same. No other info.
Not sufficient.

Using both stmnts, we know that we have B blue balls, (B + 15) Orange balls and 5B total balls.

Probability of picking an Orange ball = (B + 15) / 5B

Now as the value of B varies, the probability changes. Say B = 10, probability = 25/50
Say B = 20, probability = 35/100
and so on...

Answer (E)

St1:
P(B) = 1/4[P(G)+P(O)]
pb+pg+po=1
pb= 1/5
without other values, it is not sufficient


St2:
P(O) -15/total balls(T) = P(Blue)
we don't T
not sufficient


St1+st2:
p(o)= p(b)+15/T
replacing p(b): p(O)= 1/5+15/T

Total number of balls are not given = Insufficient

hence E