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

Solución:

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 $N = 144.$

2014 AIME I Problema 2

Solución:

El Problema 2 en otros años