Probability Test

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Probability Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

A random variable X taking values 0, 1, 2, ..., n is said to have a binomial distribution with parameters n and p, if its probability distribution is given by

A
P (X = r) = Cⁿ_r−2p^rq^n−r

B
P (X = r) = Cⁿ_rp^rq^n−r

C
P (X = r) = Cⁿ_rp^2rq^n−r

D
P (X = r) = Cⁿ_rp^rq^n−r-2

Solution

A random variable X taking values 0, 1, 2, ..., n is said to have a binomial distribution with parameters n and p, if its probability distribution is given by :P (X = r) = Cⁿ_rp^rq^n−r
Question 2
1 / -0

An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable ?

A
X = 0, 1, 2; yes

B
X = 2, 3, 5; yes

C
X = 2, 3, 4; no

D
X = 2, 1, 3; yes

Solution

An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls.therefore the possible values of X = 0 , 1 ,2 . Yes , X is a random variable.
Question 3
1 / -0

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?

A
X = 6, 4, 2, 0

B
X = 6, 4, 2, 1

C
X = 6, 3, 2, 1

D
X = 6, 5, 2, 0

Solution

Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. When a coin is tossed 6 times , we have 64 outcomes which consists of : (i) 6 heads and 0 tails (ii) 5 heads and 1 tail (iii) 4 heads and 2 tails (iv) 3 heads and 3 tails (v) 2 heads and 4 tails (vi) 1 head and 5 tails (vii) 0 head and 6 tails . Let X represents the difference between the number of heads and tails.

(i) ⇒ X = 6 – 0 = 6

(ii) ⇒ X = 5 – 1 = 4

(iii) ⇒ X = 4 – 2 = 2

(iv) ⇒ X = 3 – 3 = 0

(v) ⇒ X = 4 – 2 = 2

(vi) ⇒ X = 5 – 1 = 4

(vii) ⇒ X = 6 – 0 = 6 .

Therefore , X = 6 , 4 , 2, 0 .