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)=?](https://tex.z-dn.net/?f=P%28A%29%3D0.2%2C~~P%28B%29%3D0.3%2C~~P%28A%5Ccup%20B%29%3D%3F)
Since A and B are disjoint, so
![A\cap B=\phi~~~~~\Rightarrow P(A\cap B)=0.](https://tex.z-dn.net/?f=A%5Ccap%20B%3D%5Cphi~~~~~%5CRightarrow%20P%28A%5Ccap%20B%29%3D0.)
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.](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%3DP%28A%29%2BP%28B%29-P%28A%5Ccap%20B%29%3D0.2%2B0.3-0%3D0.50.)
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}.](https://tex.z-dn.net/?f=P%28A%29%3D%5Cdfrac%7Bn%28A%29%7D%7Bn%28S%29%7D%3D%5Cdfrac%7B3%7D%7B6%7D%3D%5Cdfrac%7B1%7D%7B2%7D.)
Thus, the correct option is (D) ![\dfrac{1}{2}.](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7D.)
(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.](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7B7%7D%7B8%7D%5Ctimes%20%5Cdfrac%7B6%7D%7B7%7D%5Ctimes%20%5Cdfrac%7B5%7D%7B6%7D%3D%5Cdfrac%7B5%7D%7B8%7D%3D0.625.)
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) =?](https://tex.z-dn.net/?f=P%28A%29%3D0.2%2C~~P%28B%29%3D0.3%2C~~~P%28A%5Ccup%20B%29%20%3D%3F)
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.](https://tex.z-dn.net/?f=P%28A%5Ccap%20B%29%3DP%28A%29%5Ctimes%20P%28B%29%3D0.2%5Ctimes%200.3%3D0.06.)
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.](https://tex.z-dn.net/?f=P%28A%5Ccup%20B%29%3DP%28A%29%2BP%28B%29-P%28A%5Ccap%20B%29%3D0.2%2B0.3-0.06%3D0.5-0.06%3D0.44.)
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.