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Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die
Given: A die is tossed two times.
When a die is tossed two times then the number of observations will be (6 × 6) = 36.
Now, let X is a random variable which represents the success.
(i) Here success is given as the number greater than 4.
Now
P(X = 0) = P(number ≤ 4 in both tosses)
P(X = 1) = P(number ≤ 4 in first toss and number ≥ 4 in second case) + P(number ≥ 4 in first toss and number ≤ 4 in second case) is:
P(X = 2) = P(number ≥ 4 in both tosses)
Hence, the required probability distribution is,
X | 0 | 1 | 2 |
P(X) | 4/9 | 4/9 | 1/9 |
(ii) Here success is given as six appears on at least one die.
Now
P(X = 0) = P(six does not appear on any of die)
P(X = 1) = P(six appears atleast once of the die)
Hence, the required probability distribution is,
X | 0 | 1 |
P(X) | 25/36 | 5/18 |
A random variable X has the following probability distribution:
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
P(X) | 0 | k | 2k | 2k | 3k | K2 | 2 K2 | 7 K2 + k |
Determine
(i) k (ii) P(X < 3)
(iii) P(X > 6) (iv) P(0 < X < 3)