Example Question - without replacement
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Calculating Probabilities with Replacement and Without Replacement
Claro, vamos a resolver paso a paso cada una de las probabilidades indicadas en la imagen: a) La probabilidad de que las dos bolas sean rojas, con reposición. Hay 4 bolas rojas de 24. Al ser con reposición, la probabilidad de sacar una bola roja en cada intento es la misma. Por lo tanto, la probabilidad de sacar una bola roja es \( \frac{4}{24} \) o \( \frac{1}{6} \). La probabilidad de sacar una bola roja en el primer intento y luego otra vez en el segundo intento (ya que se reemplaza la primera bola antes de sacar la segunda) es: \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \). b) La probabilidad de que las dos bolas sean azules, sin reposición. En este caso, hay un total de 7 bolas azules y 24 bolas en total. La probabilidad de sacar una bola azul en la primera extracción es \( \frac{7}{24} \). Tras sacar una bola azul, solo quedan 6 bolas azules y 23 bolas en total. Por tanto, la probabilidad de sacar otra bola azul es \( \frac{6}{23} \). Multiplicamos ambas probabilidades para obtener la probabilidad final de ambos eventos: \( \frac{7}{24} \times \frac{6}{23} = \frac{42}{552} = \frac{7}{92} \). c) La probabilidad de que la primera bola sea azul y la segunda verde, con reposición. La probabilidad de sacar una bola azul es \( \frac{7}{24} \), y la de sacar una bola verde después de reponer la primera bola es \( \frac{8}{24} \) o \( \frac{1}{3} \), ya que la composición de la bolsa sigue siendo la misma. La probabilidad conjunta es: \( \frac{7}{24} \times \frac{1}{3} = \frac{7}{72} \). d) La probabilidad de que la primera bola sea roja y la segunda negra, sin reposición. La probabilidad de sacar una bola roja primero es \( \frac{4}{24} \) o \( \frac{1}{6} \). Una vez que una bola roja ha sido sacada, nos quedan 23 bolas, de las cuales, según la información, \( 24 - (4 + 7 + 8) = 5 \) son negras. Entonces, la probabilidad de sacar una bola negra después de haber sacado una roja es \( \frac{5}{23} \). Multiplicamos ambas probabilidades para obtener la probabilidad final: \( \frac{1}{6} \times \frac{5}{23} = \frac{5}{138} \). Estas son las probabilidades requeridas para cada uno de los escenarios dados.
Finding Probability of Picking Even Numbers from a Set of Cards
This problem is about finding the probability of picking an even number from a set of cards labeled with numbers from 1 to 7, and then, without replacing the first card, picking another even number. Let's determine the probability step by step. 1. The probability of picking an even number (2, 4, 6) on the first draw: There are 3 even numbers out of 7 total numbers, so the probability is \( \frac{3}{7} \). 2. The probability of picking another even number on the second draw: After one even card is removed, there are now 2 even numbers remaining, and only 6 cards in total to choose from. So the probability for the second draw is \( \frac{2}{6} \) which simplifies to \( \frac{1}{3} \). Now we multiply the probabilities of both events happening in sequence, which are independent in this context: \( \frac{3}{7} \times \frac{1}{3} = \frac{3}{21} \), which simplifies to \( \frac{1}{7} \). So, the probability of drawing an even number and then another even number without replacement is \( \frac{1}{7} \).
Calculating Dependent Probability of Sequential Events
To solve this problem, we need to calculate the probability of two events happening one after the other without replacement, which is a dependent probability scenario. The two events are: 1. Picking a card with the number 1 on it. 2. Picking a card with an even number on it (without replacing the first card). First, let's calculate the probability of the first event: There are 3 cards, and only one card has the number 1 on it. So, the probability of picking the number 1 is: P(1) = 1/3 Now, assuming you picked the card with the number 1 on it, there are now 2 cards left. The second event is picking an even number, and among the remaining 2 cards (since the first card is not replaced), there is only one even number (which is number 2). So, the probability of picking an even number after picking number 1 is: P(even | 1) = 1/2 To find the overall probability of both events happening in succession, we multiply the probabilities of each individual event: P(1 and then an even) = P(1) * P(even | 1) = (1/3) * (1/2) = 1/6 Therefore, the probability of picking a 1 and then picking an even number is 1/6.
Probability of Successive Even Number Selections without Replacement
The image shows three cards with the numbers 7, 8, and 9. To find the probability of picking an even number and then picking an even number again (without replacing the first card), we follow these steps: 1. Determine the probability of picking an even number on the first draw. There is only one even number (8), and there are three cards total, so the probability is \( \frac{1}{3} \). 2. After picking an even number the first time, there are now only two cards left, and since we've taken out the 8 (the only even number), there are no even numbers left. Therefore, the probability of picking an even number on the second draw is \( \frac{0}{2} \) because there are zero even numbers out of the two remaining cards. The probability of both independent events happening is the product of their individual probabilities: \( \frac{1}{3} \times \frac{0}{2} = 0 \) Thus, the probability of picking an even number and then picking an even number again without replacement is 0. This makes sense intuitively, because once we've picked the only even number (8), it's not possible to pick another even number since it won't be replaced.
Impossibility of Successive Even Number Picks
The image displays three cards with the numbers 7, 8, and 9 on them and asks for the probability of picking an even number and then picking an even number again. However, given these three cards, this task isn't actually possible, because there is only one even number (8) among them. Once you pick the even number 8, there are only two odd numbers left (7 and 9), which means you can't pick another even number since no other even numbers are available. For illustrative purposes, let's calculate the probability as if these were the only cards and you are picking without replacement: 1. Probability of picking an even number (the 8) on the first draw: \[ P(\text{even first}) = \frac{\text{Number of even cards}}{\text{Total number of cards}} = \frac{1}{3} \] 2. Since this is without replacement, we now have two cards left, both of which are odd. The probability of picking another even card from the remaining two cards is: \[ P(\text{even second} | \text{even first}) = \frac{\text{Number of remaining even cards}}{\text{Total remaining cards}} = \frac{0}{2} \] Since it is not possible to select another even card after the first even card has been selected, the overall probability of picking an even number and then another even number is zero. The probability of an impossible event is always 0.
Calculating Probability of Picking Numbers Without Replacement
To solve this problem, we need to determine the probability of picking a 6 first and then a 5 without replacing the first number. The total number of possible numbers to choose from initially is 8 (since we see numbers 2 through 9). The probability of picking a 6 first is: 1 out of 8 (since there is only one 6) After picking a 6, there are now 7 numbers left to choose from. The probability of picking a 5 next is: 1 out of 7 (since there is only one 5) To find the combined probability of both events occurring (picking a 6 and then a 5), we multiply the probabilities of each individual event: \( (1/8) * (1/7) = 1/56 \) As a percentage, \( (1/56) \) can be calculated as follows: \( (1/56) * 100 \approx 1.786\% \) Rounding to the nearest tenth, the answer is 1.8%.
Probability of Drawing Socks from a Set Without Replacement
Este es un problema de probabilidad que involucra el sacar elementos sin reemplazo de un conjunto. Vamos a resolver cada inciso paso a paso. a) Probabilidad de que los dos calcetines sean negros. Para el primer calcetín, hay 14 calcetines negros y un total de 24 calcetines (14 + 10). Entonces, la probabilidad de que el primer calcetín sea negro es: P(primer negro) = 14/24 Si el primer calcetín es negro, hay ahora 13 calcetines negros y un total de 23 calcetines en el cajón. Entonces, la probabilidad de que el segundo calcetín también sea negro es: P(segundo negro | primer negro) = 13/23 La probabilidad conjunta de que ambos eventos ocurran es el producto de las probabilidades, ya que los eventos son dependientes (el resultado del segundo depende del primero): P(ambos negros) = P(primer negro) * P(segundo negro | primer negro) P(ambos negros) = (14/24) * (13/23) P(ambos negros) = (14 * 13) / (24 * 23) P(ambos negros) = 182 / 552 P(ambos negros) ≈ 0.3297 b) Probabilidad de que uno sea de cada color. Para que uno sea negro y el otro sea blanco, puede ocurrir de dos formas: primero negro y segundo blanco, o primero blanco y segundo negro. Primero negro y segundo blanco: P(primer negro) = 14/24 P(segundo blanco | primer negro) = 10/23 Segundo caso, primero blanco y segundo negro: P(primer blanco) = 10/24 P(segundo negro | primer blanco) = 14/23 Sumamos las probabilidades de ambos casos: P(uno de cada color) = P(primer negro) * P(segundo blanco | primer negro) + P(primer blanco) * P(segundo negro | primer blanco) P(uno de cada color) = (14/24) * (10/23) + (10/24) * (14/23) P(uno de cada color) = (140/552) + (140/552) P(uno de cada color) = 280 / 552 P(uno de cada color) ≈ 0.5072 c) Probabilidad de que ambos sean blancos. Similar al caso de los negros, tenemos: P(primer blanco) = 10/24 P(segundo blanco | primer blanco) = 9/23 P(ambos blancos) = P(primer blanco) * P(segundo blanco | primer blanco) P(ambos blancos) = (10/24) * (9/23) P(ambos blancos) = (10 * 9) / (24 * 23) P(ambos blancos) = 90 / 552 P(ambos blancos) ≈ 0.1630 d) Probabilidad de NO sacar dos calcetines blancos. La probabilidad de NO sacar dos calcetines blancos es el complemento del evento de sacar ambos blancos: P(NO ambos blancos) = 1 - P(ambos blancos) P(NO ambos blancos) = 1 - (90 / 552) P(NO ambos blancos) = 1 - 0.1630 P(NO ambos blancos) ≈ 0.8370 Estos cálculos proporcionan las probabilidades pedidas para cada situación.
Probability of Picking Two 6s from a Set of Cards
The image shows a set of four cards numbered 6, 7, 8, and 9. You are picking a card at random, and without replacing the first card, you pick a second card. The question asks what the probability is of picking a 6 and then picking another 6. Since there is only one 6 in the set of four cards, you can only pick the 6 once. This means it is impossible to pick a 6 and then pick another 6 when picking two cards without replacement. The probability, therefore, is 0%, as the event cannot occur under the given conditions.
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