Question

Now say we have an infinite sequence of independent random variables Xi (that is to say X1;X2;X3; ::::) with density f(x) stated earlier. c. What is the probability that the first random variable/trial to be greater than 2 is on the 10 trial (first 9 trials are less than 2 and the 10 trial is greater than 2)

Answer 1

Answer:

a) The value of c that will make the density function a probability density function = 0.0430

b) P(X=2) = 0.387

P(X > 2) = 0.35475 = 0.355 to 3 s.f

c) The probability that the first random variable/trial to be greater than 2 is on the 10 trial (first 9 trials are less than 2 and the 10 trial is greater than 2) = 0.0000018125 = (1.8125 × 10⁻⁶)

The probability of getting a value greater than 2 on the 10th trial and getting a Value NOT greater than 2 on first 9 trials = 0.006878

d) 0.9806

Step-by-step explanation:

Continuous random variable X with density function f(x) = c (1 + x³) where c = constant

with a support SX = [0,3]

a) Value of c that will make the density function a valid probability density function.

A valid probability density function sums up to give 1 over the interval of its sample space. That is, the sum of the probabilities over its sample space equals 1.

P(U) = ∫³₀ f(x) dx = 1

∫³₀ c (1 + x³) dx = 1

c ∫³₀ (1 + x³) dx = 1

c [x + (x⁴/4)]³₀ = 1

c [3 + (3⁴/4)] = 1

23.25c = 1

c = (1/23.25)

c = 0.0430

The value of c that will make the density function a probability density function = 0.0430

b) Probability that X = 2

f(x) = 0.043 (1 + x³)

P(X=2) = f(2) = 0.043 (1 + 2³) = 0.043 (1 + 8)

P(X=2) = 0.043 × 9 = 0.387

Probability that X is greater than 2 = P(X > 2)

P(2 ≤ X ≤ 3) = ∫³₂ f(x) dx

P(2 ≤ X ≤ 3) = ∫³₂ 0.043 (1 + x³) dx

= 0.043 ∫³₂ (1 + x³) dx

= 0.043 [x + (x⁴/4)]³₂

= 0.043 {[3 + (3⁴/4)] - [2 + (2⁴/4)]}

= 0.043 {23.25 - 6}

= 0.043 × 17.25

P(2 ≤ X ≤ 3) = P(X ≥ 2) = 0.74175

P(X > 2) = P(X ≥ 2) - P(X=2)

= 0.74175 - 0.387 = 0.35475

c) the probability that the first random variable/trial to be greater than 2 is on the 10 trial (first 9 trials are less than 2 and the 10 trial is greater than 2)

P(X>2) = 0.35475

P(X<2) = 1 - P(X ≥ 2) = 1 - 0.74175 = 0.25825

The required probability = [P(X<2)]⁹ × [P(X>2)] = (0.25825)⁹ × (0.35475) = 0.0000018125 = (1.8125 × 10⁻⁶)

Although, this probability could also be interpreted as the probability of getting a Value greater than 2 on the 10th trial and getting a value NOT greater than 2 on first 9 trials.

P(X>2) = 0.35475

Probability of X not greater than 2 = P(X<2) + P(X=2) = 1 - 0.35475 = 0.64525

Required probability = [P(X≤2)]⁹ × P(X>2) = (0.64525)⁹ × 0.35475 = 0.006878

d) Probability that it will take less than 10 random variables/trials before we see a trial that is greater than 2?

This is a sum of probabilities from getting a trial greater than 2 on the first attempt to getting it on the 9th attempt (less than 10 trials)

P(X>2) = 0.35475

Probability of X not greater than 2 = P(X<2) + P(X=2) = 1 - 0.35475 = 0.64525

Required probability = [0.35475 + (0.64525)(0.35475) + (0.64525)²(0.35475) + (0.64525)³(0.35475) + (0.64525)⁴(0.35475) + (0.64525)⁵(0.35475) + (0.64525)⁶(0.35475) + (0.64525)⁷(0.35475) + (0.64525)⁸(0.35475)] = 0.980611147 = 0.9806

Hope this Helps!!!