A coin is constructd such dt tail cn appear 4 time as likely as d head appear. Dis coin is tossd wit a fair die. Define X= no of head on coin + no of head on die.
1. Wryt out d sample space
2. Whts d prob. Of X is 14.
3. Pr(X<=10)
4. Pr(X>=14)
To solve this problem, we need to first define the sample space, which is the set of all possible outcomes. Then, we can calculate the probabilities of certain events occurring.
1. Sample Space:
Since the coin is constructed such that tails appear four times as likely as heads, we can assign probabilities of 4/5 to tails and 1/5 to heads. The die is fair, so each side has a probability of 1/6. Now, we can write out the sample space:
Sample Space = {(H,1), (H,2), (H,3), (H,4), (H,5), (H,6), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
Here, the first component represents the outcome of the coin toss (H for head, T for tail), and the second component represents the outcome of the die roll (1, 2, 3, 4, 5, 6).
2. Probability of X is 14:
To find the probability of X being 14, we need to identify all the outcomes from the sample space that satisfy this condition (i.e., the sum of the number of heads on the coin and the number of heads on the die is 14). In this case, there is only one outcome that satisfies this condition, which is (T, 6). Thus, the probability of X being 14 is:
P(X=14) = P(T, 6) = (4/5) * (1/6) = 4/30 = 2/15
3. Probability of X less than or equal to 10:
To find the probability of X being less than or equal to 10, we need to identify all the outcomes from the sample space that satisfy this condition. In this case, the possible outcomes are:
{(H,1), (H,2), (H,3), (H,4), (H,5), (T,1), (T,2), (T,3), (T,4), (T,5)}
These are the outcomes where the sum of the number of heads on the coin and the number of heads on the die is less than or equal to 10. Since there are 10 outcomes in total, the probability of X being less than or equal to 10 is:
P(X<=10) = 10/12 = 5/6
4. Probability of X greater than or equal to 14:
To find the probability of X being greater than or equal to 14, we need to identify all the outcomes from the sample space that satisfy this condition. In this case, there are no outcomes that satisfy this condition, as the maximum value for X is 13 (when the coin shows heads and the die shows 6). Therefore, the probability of X being greater than or equal to 14 is 0.
P(X>=14) = 0