Can anyone please help me understand these math problems.

#4. How many three-symbol codes (letter-number-number)can be made from the letters S, P, Y, and two-digits from the set of (0, 1, 2,..., 9) without repetition?
#16. A pair of dice are rolled and the sum on their upturned faces is recorded. What is the probability that the sum showing is 8, given that one die is showing a 5?
#20. A box contains two black balls and three gold balls. Two balls are randomly drawn in succession fro the box. a.) If there is no replacement, what is the probability that both balls are black? b) If there is replacement before the second draw, what is the probability that both balls are blacks?
#26. Jack is taking a four-item true-false test. He has no knowledge about the subject of the test and decides to flip a coin to answer the items. What is the probability that he receives a perfect score? What is the probability on a 10-item test?
Please help me.

We do not do your work for you. Once you have answered your questions, we will be happy to give you feedback on your work. Although it might require more time and effort, you will learn more if you do your own work. Isn't that why you go to school?

However, I will give you a start.

20. With replacement P = # in category/total. Without replacement, category and total are reduced by one each time. Probability of two more events all occurring is found by multiplying the probability of the individual events.

15X1

Of course, I'll be happy to help you understand these math problems. Let's solve them one by one:

#4. To solve this problem, we need to count the number of three-symbol codes that can be made by selecting one letter from the set {S, P, Y} and two digits from the set {0, 1, 2, ..., 9} without repetition.

To count the number of ways we can select one letter, we have 3 options. After selecting a letter, we will have 10 digits remaining, out of which we need to select two without repetition. The number of ways to choose two digits without repetition from a set of 10 is denoted as "10 choose 2" or C(10, 2), which is calculated as follows:

C(10, 2) = 10! / (2! * (10-2)!) = (10 * 9) / (2 * 1) = 45

Since we have 3 options for selecting a letter and 45 options for selecting two digits, the total number of three-symbol codes is given by 3 multiplied by 45:

Total number of three-symbol codes = 3 * 45 = 135

Therefore, there are 135 different three-symbol codes that can be made from the letters S, P, Y, and two digits without repetition.

#16. In this problem, we need to find the probability that the sum of two dice is 8, given that one die is showing a 5.

We know that one of the dice is showing a 5, so we only need to consider the remaining die. To get a sum of 8, the other die must show a 3.

Since each die has 6 possible outcomes (numbers 1 to 6), the total number of possible outcomes when one die shows a 5 is 6.

Out of these 6 possible outcomes, only 1 outcome is favorable (when the second die shows a 3).

Therefore, the probability of obtaining a sum of 8, given that one die shows a 5, is 1/6.

#20. a) In this problem, we are drawing two balls without replacement from a box containing 2 black balls and 3 gold balls. We need to find the probability that both balls drawn are black.

Since we are not replacing the balls after each draw and the order matters, the probability will be different for each draw.

In the first draw, there are a total of 5 balls in the box (2 black and 3 gold). The probability of drawing a black ball is 2/5.

After one black ball has been drawn, there are 4 balls remaining in the box (1 black and 3 gold). For the second draw, the probability of drawing another black ball is 1/4.

To find the probability of both events happening, we multiply the probabilities of each event:

Probability of drawing a black ball in the first draw = 2/5
Probability of drawing a black ball in the second draw after one black ball has been drawn = 1/4

Probability of drawing two black balls without replacement = (2/5) * (1/4) = 1/10

Therefore, the probability that both balls drawn are black, without replacement, is 1/10.

b) In this case, we are replacing the ball after each draw. So the probability of drawing a black ball remains the same for both draws.

The probability of drawing a black ball in each draw is 2/5.

To find the probability of both events happening, we multiply the probabilities of each event:

Probability of drawing a black ball in the first draw = 2/5
Probability of drawing a black ball in the second draw (with replacement) = 2/5

Probability of drawing two black balls with replacement = (2/5) * (2/5) = 4/25

Therefore, the probability that both balls drawn are black, with replacement, is 4/25.

#26. In this problem, Jack is taking a four-item true-false test. Since he has no knowledge about the subject, he decides to flip a coin to answer each item. We need to find the probability that he receives a perfect score.

For each item, Jack has two possible outcomes: heads or tails (true or false). Since there are 4 items, the total number of possible outcomes is 2^4 = 16.

Out of these 16 possible outcomes, only 1 outcome will result in a perfect score (HHHH, where H represents heads and means Jack answers all items correctly).

Therefore, the probability that he receives a perfect score on a four-item test is 1/16.

For a 10-item test, the same logic applies. Jack has two possible outcomes for each item, so the total number of possible outcomes is 2^10 = 1024.

Out of these 1024 possible outcomes, only 1 outcome will result in a perfect score (HHHHHHHHHH, where H represents heads and means Jack answers all items correctly).

Therefore, the probability that he receives a perfect score on a ten-item test is 1/1024.

I hope this helps you understand these math problems. If you have any further questions, feel free to ask!