A random variable X has the following probability distribution:

Determine:

(i) K (ii) P (X < 3)

(iii) P (X > 6) (iv) P (0 < X < 3)

OR

Find the probability of throwing at most 2 sixes in 6 throws of a single die

To find the value of k we will be using the very basic idea of probability.

Note: We know that the sum of the probabilities of all random variables taken from a given sample space is equal to 1.

i) Finding k

∴ = 1

∴ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1

⇒ 10k² + 9k – 1 = 0

⇒10k² + 10k - k – 1 = 0

⇒10k(k+1) – (k+1) = 0

⇒ (10k-1)(k+1) = 0

∴ k = 1/10 = 0.1 or k = -1

∵ k represents probability of an event. Hence 0≤P(X)≤1

∴ k = 1/10 = 0.1

ii) P(X < 3)

We know that: P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

∴ P(X < 3) = 0 + k + 2k {referring to given table}

⇒ P(X < 3) = 3k = 3*0.1 = 0.3

∴ P(X < 3) = 0.3

iii) P(X > 6)

We know that: P(X < 6) = P(X = 7)

∴ P(X > 6) = 7k² + k {referring to given table}

⇒ P(X > 6) = 7(0.1)² + 0.1 = 0.07 + 0.1 = 0.17

∴ P(X > 6) = 0.17

iv) P(0< X < 3)

We know that: P(0< X < 3) = P(X = 1) + P(X = 2)

∴ P(X < 3) = k + 2k {referring to given table}

⇒ P(X < 3) = 3k = 3*0.1 = 0.3

∴ P(0< X < 3) = 0.3

OR

As we know that repeated throws of a die are Bernoulli trials.

Let X denotes the number of sixes in 6 throws of die.

∴ X has the binomial distribution with n= 6

p = probability of success(or probability of getting 6) = 1/6

and q = probability of failure = probability of not getting 6 = 5/6

∴ Probability of getting at most 2 sixes in 6 throws = P(X<3)

P(X = 0) + P(X = 1) + P(X = 2) = ⁶C₀.p⁰q⁶ + ⁶C₁p.q⁵ + ⁶C₂p²q³

⇒ P(X < 3) =

⇒ P(X < 3) =

⇒ P(X < 3) =

⇒ P(X < 3) =

∴ The probability of getting at most 2 sixes in 6 throws of dice = 0.9377