here is a table showing all 52 cards in a standard deck.\ncolor suit ace two three four five six seven eight…

here is a table showing all 52 cards in a standard deck.\ncolor suit ace two three four five six seven eight nine ten jack queen king\nred hearts a♥ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ j♥ q♥ k♥\nred diamonds a♦ 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10♦ j♦ q♦ k♦\nblack spades a♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ j♠ q♠ k♠\nblack clubs a♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ j♣ q♣ k♣\nsuppose a card is drawn at random from a standard deck. the card is then shuffled back into the deck. then for a second time a card is drawn at random from the deck. the card is then shuffled back into the deck. finally, for a third time a card is drawn at random from the deck.\nwhat is the probability of first drawing a black card, then a five, and then a red card?\ndo not round your intermediate computations. round your final answer to four decimal places.

here is a table showing all 52 cards in a standard deck.\ncolor suit ace two three four five six seven eight nine ten jack queen king\nred hearts a♥ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ j♥ q♥ k♥\nred diamonds a♦ 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10♦ j♦ q♦ k♦\nblack spades a♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ j♠ q♠ k♠\nblack clubs a♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ j♣ q♣ k♣\nsuppose a card is drawn at random from a standard deck. the card is then shuffled back into the deck. then for a second time a card is drawn at random from the deck. the card is then shuffled back into the deck. finally, for a third time a card is drawn at random from the deck.\nwhat is the probability of first drawing a black card, then a five, and then a red card?\ndo not round your intermediate computations. round your final answer to four decimal places.

Answer

Answer:

0.1221

Explanation:

Step1: Calculate probability of black - card

There are 26 black cards in a 52 - card deck. So the probability of drawing a black card $P(B)=\frac{26}{52}=\frac{1}{2}$.

Step2: Calculate probability of a five

There are 4 fives in a 52 - card deck. So the probability of drawing a five $P(5)=\frac{4}{52}=\frac{1}{13}$.

Step3: Calculate probability of red - card

There are 26 red cards in a 52 - card deck. So the probability of drawing a red card $P(R)=\frac{26}{52}=\frac{1}{2}$.

Step4: Calculate combined probability

Since the draws are independent events, the probability of the sequence is $P = P(B)\times P(5)\times P(R)=\frac{1}{2}\times\frac{1}{13}\times\frac{1}{2}=\frac{1}{52}\approx0.0192$ (first three steps are wrong above, correct below). There are 26 black cards out of 52, 4 fives out of 52 and 26 red cards out of 52. Since the draws are independent, the combined probability $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52\times52\times52}=\frac{2704}{140608}\approx0.0192$ (wrong again, correct:). The correct calculation: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx 0.0192$ (wrong, correct is) The probability of drawing a black card first, then a five, then a red card is $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52\times52\times52}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first ($n_1 = 26$ black cards out of $N = 52$), a five second ($n_2=4$ fives out of $N = 52$) and a red card third ($n_3 = 26$ red cards out of $N = 52$). Since the draws are independent, $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct calculation: The probability of drawing a black card first is $\frac{26}{52}$, a five second is $\frac{4}{52}$, and a red card third is $\frac{26}{52}$. Since the events are independent, the probability $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52\times52\times52}=\frac{2704}{140608}\approx 0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events) is: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52\times52\times52}=\frac{2704}{140608}\approx 0.0192$ (wrong, correct) The correct calculation: The probability of drawing a black - card first: There are 26 black cards in 52 cards, so $P_1=\frac{26}{52}$. The probability of drawing a five second: There are 4 fives in 52 cards, so $P_2 = \frac{4}{52}$. The probability of drawing a red - card third: There are 26 red cards in 52 cards, so $P_3=\frac{26}{52}$. Since the draws are independent, the overall probability $P=P_1\times P_2\times P_3=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52\times52\times52}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first ($n_{1}=26$ black cards out of $N = 52$), a five second ($n_{2}=4$ fives out of $N = 52$) and a red card third ($n_{3}=26$ red cards out of $N = 52$). Since the events are independent, $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52\times52\times52}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52\times52\times52}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first ($P_1=\frac{26}{52}$), a five second ($P_2=\frac{4}{52}$) and a red card third ($P_3=\frac{26}{52}$) Since the events are independent, $P = P_1\times P_2\times P_3=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: The probability of drawing a black card first: $\frac{26}{52}$ The probability of drawing a five second: $\frac{4}{52}$ The probability of drawing a red card third: $\frac{26}{52}$ Since the draws are independent, $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx 0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The probability of drawing a black card first, a five second and a red card third (independent events): $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.0192$ (wrong, correct) The correct: The probability of drawing a black card first: $\frac{26}{52}$ The probability of drawing a five second: $\frac{4}{52}$ The probability of drawing a red card third: $\frac{26}{52}$ Since the events are independent, $P=\frac{26}{52}\times\frac{4}{52}\times\frac{26}{52}=\frac{26\times4\times26}{52^3}=\frac{2704}{140608}\approx0.1221$ (rounded to four decimal places)