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  1. 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

  1. 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

  1. 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

  1. 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}

  1. 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

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

3

4

6

8

9

Verified Answer :

Answer:  The calculations are below.

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

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

\(P(A)=0.2,~~P(B)=0.3,~~P(A\cup B)=?\)

Since A and B are disjoint, so

\(A\cap B=\phi~~~~~\Rightarrow P(A\cap B)=0.\)

From the law of probability, we have

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.2+0.3-0=0.50.\)

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

\(P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{6}=\dfrac{1}{2}.\)

Thus, the correct option is (D) \(\dfrac{1}{2}.\)

(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

\(P=\dfrac{7}{8}\times \dfrac{6}{7}\times \dfrac{5}{6}=\dfrac{5}{8}=0.625.\)

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,

\(P(A)=0.2,~~P(B)=0.3,~~~P(A\cup B) =?\)

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

\(P(A\cap B)=P(A)\times P(B)=0.2\times 0.3=0.06.\)

From the law of probability, we have

\(P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.2+0.3-0.06=0.5-0.06=0.44.\)

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.

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