3. if p (a) = 0.2 and p (b) = 0.3 and a and b are disjoint, what is p (a or b)?

0.00
0.06
0.10
0.44
0.50

4. given that you roll a fair six-sided die, what is the probability that you roll an odd number? (1 point)

0
1/6
1/3
1/2
1

5. there are 3 red marbles, 4 white marbles, and 1 green marble in a bag. marbles are drawn without replacement. what is the probability that 3 marbles can be drawn without drawing the green marble? (1 point)

0.625
0.643
0.670
0.998
0.179

6. which of the following represents the sample space for rolling a six-sided die and tossing a coin? (1 point)

{1h, 2h, 3h, 4h, 5h, 6h, 1t, 2t, 3t, 4t, 5t, 6t}
{1, 2, 3, 4, 5, 6}
{h, t}
{1, 2, 3, 4, 5, 6, h, t}

7. if p (a) = 0.2 and p (b) = 0.3 and a and b are independent but not necessarily disjoint, find p (a or b). (1 point)

0.06
0.10
0.44
0.50
cannot be determined from the information given

8. if three coins are tossed, what is the number of equally likely outcomes? (1 point)

3
4
6
8
9

 The calculations are below.

Step-by-step explanation:  The calculations are as follows:

(3) Given that for two events A and B,

Since A and B are disjoint, so

From the law of probability, we have

Thus, the correct option is (E) 0.50.

(4) Given that a fair six-sided die is rolled. We are to find the probability that an odd number is rolled.

Let, 'A' be event of rolling an odd number. So,

A = {1, 3, 5}   ⇒ n(A) = 3.

Let 'S' be the sample space for the experiment. So,

S = {1, 2, 3, 4, 5, 6}   ⇒ n(S) = 6.

Therefore, the probability of rolling an odd number is given by

Thus, the correct option is (D)

(5) Given that there are 3 red marbles, 4 white marbles, and 1 green marble in a bag and marbles are drawn without replacement.

So, the probability that 3 marbles can be drawn without drawing the green marble is given by

Thus, the correct option is (A) 0.625.

(6) The sample space for rolling a six-sided die is  {1, 2, 3, 4, 5, 6}

and the sample space for tossing a coin is {H, T}.

Therefore, the sample space for rolling a six-sided die and tossing a coin will be

S = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}.

Thus, the correct option is {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}.

(7) Given that for two events A and B,

Since A and B are independent but not necessarily disjoint, so

From the law of probability, we have

Thus, the correct option is (C) 0.44.

(8) Given that three coins are tossed.

The sample space for each of them is S = {1, 2, 3, 4, 5, 6}.

The equally likely outcomes when three coins are tossed together are

{(1, 1 , 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6)}.

So, the total number of equally likely outcomes = 6.

Thus, the correct option is (C) 6.

Hence all the questions are answered.

b.-4 units

all you   do is count how many times you move

answer: he is 272195

step-by-step explanation: all you have to do is multiple 3535 and 77 and it equals 272195