Problem Solving

BTGmoderatorLU wrote:Source: Veritas Prep

A basketball team's current roster consists of g guards and f forwards. If 2 guards join and 1 forward leaves, what is the probability that a player chosen at random is a guard?

A. \(\frac{g}{g+f}\)

B. \(\frac{g+2}{g+f}\)

C. \(\frac{g+2}{g+f+3}\)

D. \(\frac{g+2}{g+f+1}\)

E. \(\frac{g+2}{g+f+2}\)

The OA is D

The number of guards now = g + 2 and the number of forwards now = f - 1

Total number of players = g + 2 + f - 1 = g + f +1

The probability that a player chosen at random is a guard = number of guards / total number of players = (g + 2) / (g + f +1)

The correct answer: D

Hope this helps!

-Jay
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BTGmoderatorLU wrote:Source: Veritas Prep

A basketball team's current roster consists of g guards and f forwards. If 2 guards join and 1 forward leaves, what is the probability that a player chosen at random is a guard?

A. \(\frac{g}{g+f}\)

B. \(\frac{g+2}{g+f}\)

C. \(\frac{g+2}{g+f+3}\)

D. \(\frac{g+2}{g+f+1}\)

E. \(\frac{g+2}{g+f+2}\)

The OA is D

If 2 guards join and 1 forward leaves, the new number of guards is (g + 2), and the new number of forwards is (f - 1); thus, the new total is (g + 2 + f - 1). The probability that a guard is chosen is:

(g + 2)/(g + 2 + f - 1) = (g + 2)/(g + 1 + f)

Answer: D

Hi All,

We're told that a basketball team's current roster consists of G guards and F forwards - and 2 guards join and 1 forward leaves. We're asked for the probability that a player chosen at random is a guard. This question can be approached in a couple of different ways, including by TESTing VALUES.

IF... G=2 and F=3, then after 2 guards JOIN and 1 forward LEAVES, we have....
4 total guards and 2 total forwards.
The probability of randomly choosing a guard is 4/6.

Thus, we're looking for an answer that equals 4/6 when G=2 and F=3. There's only one answer that matches...

Final Answer: D

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Rich