ANYONE, PLEASE ANSWER ALL THESE QUESTIONS AND I WILL MARK HIM/HER BRAINLIEST + 57 PTS. 1.A bag contains 15 balls of which 6 are red, the probability of getting a red ball at random from the bag is
A.9/15:
B.6/15:
C.15/9:
D.15/6:

2.In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?
A.8/21:
B.2/7:
C.1/3:
D.2/3:

3.One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
A.1/13:
B.1/4:
C.9/52:
D.3/13:

Descriptive Type Questions

4.What is the probability from two throws of a dice:
(a)For getting a sum 9 (b) For the total score is a prime number:

5.Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?:

1.B

2.C

3.D

4.

a)1/9

b)5/12

5.2/5

Step-by-step explanation:

1.

Since there are 15 total outcomes of which 6 are desirable, the probability of randomly getting a red ball is 5/16 or choice B.

2.

Since there are a total of 8+7+6=21 outcomes, out of which 7 are neither red nor green, there is a 7/21=1/3 probability of that outcome, or answer choice C.

3.

Out of the 52 possible outcomes, 12 are face cards. Therefore, the probability of drawing a face card is 12/52=3/13, or answer choice D.

4.

a)

For each die, there are a total of 6 possible outcomes. This means that in total there are 6*6=36 possible combinations. Now, let's do some casework.

If the first die is a 3, then the second die must be a 6. If the first die is a 4, the second die must be a 5. If the first die is a 5, then the second die must be a 4. And if the first die is a 6, then the second die must be a 3. You cannot have any numbers less than 3, because then you would not be able to make the number 9. Therefore, there is a 4/36=1/9 chance of getting a sum of 9.

b)

All of the possible prime numbers that can be formed with the dice are 2, 3, 5, 7, and 11. For 2, there is only one possible situation where this can happen. For 3, there are 2 possible. For 5, there is 4-1, 3-2, 2-3, and 1-4, or 4 possibilities. For 7, there is 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1, or 6 total possibilities. Finally, for 11, there is only 5-6 and 6-5, or 2 possible. In total, the probability of getting a prime sum is 15/36=5/12.

5.

The only numbers which are a multiple of 3 or 5 from 1 to 20 are: 3, 5, 9, 10, 12, 15, 18, and 20. This means that out of the 20 total possible options, 8 of them are a multiple of 3 or 5, which is a total chance of 2/5.

Hope this helps!