2014 AIME I Problema 2

Question

Una urna contiene

4

bolas verdes y

6

bolas azules. Una segunda urna contiene

16

bolas verdes y

N

bolas azules. Se extrae al azar una sola bola de cada urna. La probabilidad de que ambas bolas sean del mismo color es

0.58.

Halla

N.

An urn contains

4

green balls and

6

blue balls. A second urn contains

16

green balls and

N

blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is

0.58.

Find

N.

Respuesta

Suma las probabilidades de sacar verde-verde y azul-azul

Add the probabilities of green-green and blue-blue draws

Resuelve

\frac{4}{10} \cdot \frac{16}{16+N} + \frac{6}{10} \cdot \frac{N}{16+N} = \frac{29}{50}

para

N

Solve

\frac{4}{10} \cdot \frac{16}{16+N} + \frac{6}{10} \cdot \frac{N}{16+N} = \frac{29}{50}

for

N

Ambas bolas son verdes con probabilidad

\frac{4}{10} \cdot \frac{16}{16+N},

y ambas son azules con probabilidad

\frac{6}{10} \cdot \frac{N}{16+N}.

La condición es

\frac{64 + 6N}{10(16 + N)} = \frac{29}{50}.

Al eliminar denominadores,

5(64 + 6N) = 29(16 + N),

así que

320 + 30N = 464 + 29N,

lo que da

N = 144.

Both balls are green with probability

\frac{4}{10} \cdot \frac{16}{16+N},

and both are blue with probability

\frac{6}{10} \cdot \frac{N}{16+N}.

The condition is

\frac{64 + 6N}{10(16 + N)} = \frac{29}{50}.

Clearing denominators,

5(64 + 6N) = 29(16 + N),

so

320 + 30N = 464 + 29N,

giving

N = 144.

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