Terdapat dua kantong. Kantong pertama berisi $10$ bola, terdiri atas $3$ bola merah dan $7$ bola biru. Kantong kedua berisi $10$ bola, terdiri atas $4$ bola merah dan $6$ bola biru. Satu bola diambil dari kantong pertama, dicatat, dan disimpan pada kantong kedua. Setelah dikocok, satu bola diambil dari kantong kedua dan dikembalikan ke kantong pertama. a. Buatlah diagram pohonnya. b. Tentukan peluang memperoleh $1$ bola merah dari kantong pertama dan $1$ bola biru dari kantong kedua. c. Tentukan peluang memperoleh dua bola berwarna berbeda. d. Tentukan peluang memperoleh bola merah dari kantong kedua. Tentukan peluang bahwa kantong pertama tetap berisi $3$ bola merah setelah pengambilan dari kantong kedua dan dikembalikan ke kantong pertama.

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Algebra 2

Csc, sec, and cot of special angles

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Q. Terdapat dua kantong. Kantong pertama berisi

10

bola, terdiri atas

3

bola merah dan

7

bola biru. Kantong kedua berisi

10

bola, terdiri atas

4

bola merah dan

6

bola biru. Satu bola diambil dari kantong pertama, dicatat, dan disimpan pada kantong kedua. Setelah dikocok, satu bola diambil dari kantong kedua dan dikembalikan ke kantong pertama. a. Buatlah diagram pohonnya. b. Tentukan peluang memperoleh

1

bola merah dari kantong pertama dan

1

bola biru dari kantong kedua. c. Tentukan peluang memperoleh dua bola berwarna berbeda. d. Tentukan peluang memperoleh bola merah dari kantong kedua. Tentukan peluang bahwa kantong pertama tetap berisi

3

bola merah setelah pengambilan dari kantong kedua dan dikembalikan ke kantong pertama.

Add Red Ball Second Bag: Next, we add the red ball to the second bag, which now has $5$ red balls and $6$ blue balls, making $11$ total balls. $\newline$ Probability of blue from second bag after adding red = $\frac{6}{11}$ .
Multiply Probabilities: Now, we multiply the probabilities to find the combined probability of both events happening. $\newline$ Combined probability = $(\frac{3}{10}) \times (\frac{6}{11})$ .
Calculate Combined Probability: Perform the multiplication to get the combined probability. $\newline$ Combined probability = $\frac{18}{110}.$
Calculate Probability Different Colors: Simplify the fraction by dividing both numerator and denominator by $2$ . $\newline$ Combined probability = $\frac{9}{55}$ .
Calculate Probability Red Second Bag: Now, let's calculate the probability of getting two balls of different colors. $\newline$ First, the probability of getting a red ball from the first bag and then a blue ball from the second bag is already calculated as $\frac{9}{55}$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .Multiply the probabilities to find the combined probability of drawing a blue ball from the first bag and a red ball from the second bag. $\newline$ Combined probability for blue from first and red from second = $\left(\frac{7}{10}\right) \times \left(\frac{4}{11}\right)$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .Multiply the probabilities to find the combined probability of drawing a blue ball from the first bag and a red ball from the second bag. $\newline$ Combined probability for blue from first and red from second = $\left(\frac{7}{10}\right) \times \left(\frac{4}{11}\right)$ .Perform the multiplication to get the combined probability. $\newline$ Combined probability for blue from first and red from second = $\frac{28}{110}$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .Multiply the probabilities to find the combined probability of drawing a blue ball from the first bag and a red ball from the second bag. $\newline$ Combined probability for blue from first and red from second = $\left(\frac{7}{10}\right) \times \left(\frac{4}{11}\right)$ .Perform the multiplication to get the combined probability. $\newline$ Combined probability for blue from first and red from second = $\frac{28}{110}$ .Simplify the fraction by dividing both numerator and denominator by $2$ . $\newline$ Combined probability for blue from first and red from second = $\frac{14}{55}$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .Multiply the probabilities to find the combined probability of drawing a blue ball from the first bag and a red ball from the second bag. $\newline$ Combined probability for blue from first and red from second = $\left(\frac{7}{10}\right) \times \left(\frac{4}{11}\right)$ .Perform the multiplication to get the combined probability. $\newline$ Combined probability for blue from first and red from second = $\frac{28}{110}$ .Simplify the fraction by dividing both numerator and denominator by $2$ . $\newline$ Combined probability for blue from first and red from second = $\frac{14}{55}$ .Add the probabilities of the two different color scenarios together. $\newline$ Total probability for different colors = $\left(\frac{9}{55}\right) + \left(\frac{14}{55}\right)$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .Multiply the probabilities to find the combined probability of drawing a blue ball from the first bag and a red ball from the second bag. $\newline$ Combined probability for blue from first and red from second = $\left(\frac{7}{10}\right) \times \left(\frac{4}{11}\right)$ .Perform the multiplication to get the combined probability. $\newline$ Combined probability for blue from first and red from second = $\frac{28}{110}$ .Simplify the fraction by dividing both numerator and denominator by $2$ . $\newline$ Combined probability for blue from first and red from second = $\frac{14}{55}$ .Add the probabilities of the two different color scenarios together. $\newline$ Total probability for different colors = $\left(\frac{9}{55}\right) + \left(\frac{14}{55}\right)$ .Perform the addition to get the total probability. $\newline$ Total probability for different colors = $4$ $0$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .Multiply the probabilities to find the combined probability of drawing a blue ball from the first bag and a red ball from the second bag. $\newline$ Combined probability for blue from first and red from second = $\left(\frac{7}{10}\right) \times \left(\frac{4}{11}\right)$ .Perform the multiplication to get the combined probability. $\newline$ Combined probability for blue from first and red from second = $\frac{28}{110}$ .Simplify the fraction by dividing both numerator and denominator by $2$ . $\newline$ Combined probability for blue from first and red from second = $\frac{14}{55}$ .Add the probabilities of the two different color scenarios together. $\newline$ Total probability for different colors = $\left(\frac{9}{55}\right) + \left(\frac{14}{55}\right)$ .Perform the addition to get the total probability. $\newline$ Total probability for different colors = $4$ $0$ .Finally, calculate the probability of getting a red ball from the second bag after the transfer and return. $\newline$ Since we're returning the ball to the first bag, the second bag's composition remains the same as initially, with $4$ red balls out of $4$ $2$ . $\newline$ Probability of red from second bag = $4$ $3$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .Multiply the probabilities to find the combined probability of drawing a blue ball from the first bag and a red ball from the second bag. $\newline$ Combined probability for blue from first and red from second = $\left(\frac{7}{10}\right) \times \left(\frac{4}{11}\right)$ .Perform the multiplication to get the combined probability. $\newline$ Combined probability for blue from first and red from second = $\frac{28}{110}$ .Simplify the fraction by dividing both numerator and denominator by $2$ . $\newline$ Combined probability for blue from first and red from second = $\frac{14}{55}$ .Add the probabilities of the two different color scenarios together. $\newline$ Total probability for different colors = $\left(\frac{9}{55}\right) + \left(\frac{14}{55}\right)$ .Perform the addition to get the total probability. $\newline$ Total probability for different colors = $4$ $0$ .Finally, calculate the probability of getting a red ball from the second bag after the transfer and return. $\newline$ Since we're returning the ball to the first bag, the second bag's composition remains the same as initially, with $4$ red balls out of $4$ $2$ . $\newline$ Probability of red from second bag = $4$ $3$ .Simplify the fraction for the probability of red from the second bag. $\newline$ Probability of red from second bag = $4$ $4$ .
Calculate Probability Red First Bag: Next, calculate the probability of getting a blue ball from the first bag, which is $\frac{7}{10}$ .After transferring a blue ball to the second bag, it now has $4$ red balls and $7$ blue balls, making $11$ total balls. $\newline$ Probability of red from second bag after adding blue = $\frac{4}{11}$ .Multiply the probabilities to find the combined probability of drawing a blue ball from the first bag and a red ball from the second bag. $\newline$ Combined probability for blue from first and red from second = $\left(\frac{7}{10}\right) \times \left(\frac{4}{11}\right)$ .Perform the multiplication to get the combined probability. $\newline$ Combined probability for blue from first and red from second = $\frac{28}{110}$ .Simplify the fraction by dividing both numerator and denominator by $2$ . $\newline$ Combined probability for blue from first and red from second = $\frac{14}{55}$ .Add the probabilities of the two different color scenarios together. $\newline$ Total probability for different colors = $\left(\frac{9}{55}\right) + \left(\frac{14}{55}\right)$ .Perform the addition to get the total probability. $\newline$ Total probability for different colors = $4$ $0$ .Finally, calculate the probability of getting a red ball from the second bag after the transfer and return. $\newline$ Since we're returning the ball to the first bag, the second bag's composition remains the same as initially, with $4$ red balls out of $4$ $2$ . $\newline$ Probability of red from second bag = $4$ $3$ .Simplify the fraction for the probability of red from the second bag. $\newline$ Probability of red from second bag = $4$ $4$ .Now, determine the probability that the first bag still contains $4$ $5$ red balls after the transfer and return. $\newline$ This is the same as the initial probability since the ball is returned, so the composition of the first bag is unchanged. $\newline$ Probability that first bag still has $4$ $5$ red balls = $4$ $7$ .