Question

1. Suppose X is a discrete random variable with possible values 0, 1, 2, ..., 10 and P(X = k) = (10 choose k)(1/6)^k (5/6)^(10-k). Note that (10 choose k) = 10! / (k!(10-k)!). Find each of the following probabilities: a) P(X >= 1) b) P(X > 4) c) P(3 <= X <= 5) d) P(2.5 <= X <= 2.9) e) P(X >= 0) f) P(X <= 4)

          1. Suppose X is a discrete random variable with possible values 0, 1, 2, ..., 10 and
P(X = k) = (10 choose k)(1/6)^k (5/6)^(10-k).
Note that (10 choose k) = 10! / (k!(10-k)!).
Find each of the following probabilities:
a) P(X >= 1)
b) P(X > 4)
c) P(3 <= X <= 5)
d) P(2.5 <= X <= 2.9)
e) P(X >= 0)
f) P(X <= 4)

1. Suppose X is a discrete random variable with possible values 0, 1, 2, ..., 10 and
P(X = k) = (10 choose k)(1/6)^k (5/6)^(10-k).
Note that (10 choose k) = 10! / (k!(10-k)!).
Find each of the following probabilities:
a) P(X >= 1)
b) P(X > 4)
c) P(3 <= X <= 5)
d) P(2.5 <= X <= 2.9)
e) P(X >= 0)
f) P(X <= 4)

Added by Matthew G.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Sri K

06/16/2023

Step 1

So, we need to find P(X = 2) + P(X = 10). P(X = 2) = (2 10) = 10! / (2! * (10-2)!) = 45 P(X = 10) = (10 10) = 10! / (10! * (10-10)!) = 1 So, P(X > 1) = 45 + 1 = 46. b) P(X > 4) means we want to find the probability that X takes on the value 10. So, we need to Show more…

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Suppose X is a discrete random variable with possible values 0, 1, 2, ..., 10 and P(X = k) = (10 choose k)(1/6)^k (5/6)^(10-k). Note that (10 choose k) = 10! / (k!(10-k)!). Find each of the following probabilities: a) P(X >= 1) b) P(X > 4) c) P(3 <= X <= 5) d) P(2.5 <= X <= 2.9) e) P(X >= 0) f) P(X <= 4)