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3) En una urna hay 6 bolitas blancas y 3 azules ¿Cuál es la probabilidad de sacar al azar ambas blancas sin reposición?
\nA) $\\Large \\frac{1}{2}$
\nB) $\\Large \\frac{5}{8}$
\nC) $\\Large \\frac{2}{3}$
\nD) $\\Large \\frac{5}{12}$
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\n\n [[0]] \n$P(1°$ bolita blanca$)=$ ______ \n[[1]] \n | \nPaso 1: Calcular la probabilidad de que la primera bolita sea blanca. | \n
\n [[2]] \n$P(2°$ bolita blanca$)=$ ______ \n[[3]] \n | \n\n Paso 2: Calcular la probabilidad de que la primera bolita sea blanca. Recuerda que se debe eliminar una bolita blanca (ya que no se repone la bolita al ser sacada) y del total de bolitas también se elimina una. \n\n | \n
\n\n [[0]] [[2]] \n$P(2 $ blancas sin reponer$)= $ ______ $\\cdot$ ______ $=$ \n[[1]] [[3]] \n[[4]] \n$=$ ______ \n[[5]] \n | \nPaso 3: Como ambos sucesos ocurren uno después del otro, debemos multiplicar las probabilidades del paso 1 y 2. | \n
\n [[6]] \n$P(2$ blancas sin reponer$)=$ ______ \n[[7]] \n | \nPaso 4: Se simplifica el valor obtenido en el paso 3. | \n
Por tanto, la respuesta correcta es: [[8]]
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