Example Question - drawing cards
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Probability of Drawing Cards from a Deck
<p>Let E be the event that the card drawn is not a heart.</p> <p>There are a total of 52 cards in a deck, and there are 13 cards of each suit, including hearts.</p> <p>\( P(E) = \frac{Number\ of\ non-heart\ cards}{Total\ number\ of\ cards} \)</p> <p>\( P(E) = \frac{52 - 13}{52} \)</p> <p>\( P(E) = \frac{39}{52} \)</p> <p>\( P(E) = \frac{3}{4} \)</p> <p>So, the probability that the card drawn is not a heart is \(\frac{3}{4}\).</p>
Calculating Probability of Drawing 2 Sotas and 2 Reyes from a Deck of Spanish Cards
Para resolver esta pregunta, primero debemos entender que estamos tratando con un problema de probabilidad donde queremos calcular la posibilidad de sacar 2 sotas y 2 reyes de un mazo de 48 cartas españolas. En un mazo de cartas españolas, hay 4 sotas y 4 reyes. Aquí está cómo podemos calcular la probabilidad de sacar 2 sotas y 2 reyes al sacar 4 cartas: 1. Calculamos el número de maneras en las que podemos sacar 2 sotas de las 4 que hay: C(4,2) = 4! / [2! * (4 - 2)!] = (4 * 3) / (2 * 1) = 6 maneras 2. Luego, calculamos el número de maneras en que podemos sacar 2 reyes de los 4 que hay: C(4,2) = 4! / [2! * (4 - 2)!] = (4 * 3) / (2 * 1) = 6 maneras 3. Ahora, calculamos el número total de maneras de sacar 4 cartas de un mazo de 48 cartas: C(48,4) = 48! / [4! * (48 - 4)!] = (48 * 47 * 46 * 45) / (4 * 3 * 2 * 1) = 194580 maneras 4. La probabilidad de sacar específicamente 2 sotas y 2 reyes es el producto de las maneras de sacar las sotas y los reyes dividido por el número total de maneras de sacar 4 cartas: Probabilidad = (C(4,2) * C(4,2)) / C(48,4) Probabilidad = (6 * 6) / 194580 Probabilidad = 36 / 194580 Probabilidad ≈ 0.000185 Así que la probabilidad de sacar 2 sotas y 2 reyes al sacar 4 cartas de un mazo de 48 cartas españolas es aproximadamente 0,000185 o 0,0185%.
Calculating Probability of Drawing Cards from a Set
The question displayed in the image asks for the probability of picking an even number first and then picking an 8 when drawing cards labeled with the numbers 7, 8, and 9. Since the problem states that the card is put back before the second draw, the events are independent. To find the probability of both events happening, we calculate the probability of each event separately and then multiply them together. The probability of drawing an even number (which could only be the 8 in this case) on the first draw is 1 out of 3, since there is one even number and three cards overall. The probability of drawing an 8 on the second draw is again 1 out of 3, because after replacing the card, all three cards are available for selection again. Therefore, the combined probability is: Probability of even number (8) then an 8 = (1/3) * (1/3) = 1/9 The answer, as a simplified fraction, is 1/9.
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