CS 2203 Statistics and Probability Theory: Conditional Probability Example

The probability that the husband watches the football match given that his wife is also watching it is 0.75.

The probability of getting an odd number in the first toss and an even number in the second toss is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.

The sample space for the experiment of tossing a die twice is the set of all possible outcomes, which is ${(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}$.

<p>A deterministic event is an event where the outcome is certain and can be predicted with certainty, while a non-deterministic event is an event where the outcome is uncertain and cannot be predicted with certainty.</p> Signup and view all the answers

<p>The probability of a non-deterministic event can be calculated using the formula $P(A) = \frac{n(A)}{n(S)}$, where $A$ is the event, $n(A)$ is the number of favorable outcomes for event $A$, and $n(S)$ is the total number of possible outcomes in the sample space $S$.</p> Signup and view all the answers

<p>The probability of the husband watching the football match is 0.80, while the probability of the wife watching the football match is 0.65. The difference between these two probabilities is 0.80 - 0.65 = 0.15.</p> Signup and view all the answers

<p>The probability that both the husband and wife watch the football match is 0.60.</p> Signup and view all the answers

<p>The probability of the husband watching the match and the probability of the wife watching the match are not independent events, as the given information states that the probability that they both watch the match is 0.60, which is less than the product of their individual probabilities (0.80 * 0.65 = 0.52).</p> Signup and view all the answers

<p>The sample space for the experiment of tossing a coin twice is the set of all possible outcomes, which is ${(H,H), (H,T), (T,H), (T,T)}$.</p> Signup and view all the answers

<p>The probability of getting a sum of 7 when tossing two dice is $\frac{6}{36} = \frac{1}{6}$.</p> Signup and view all the answers