Answer :

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 + k^{2} + 2k^{2} + 7k^{2} + k = 1

⇒ 10k^{2} + 9k – 1 = 0

⇒10k^{2} + 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) = 7k^{2} + 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) = ^{6}C_{0}.p^{0}q^{6} + ^{6}C_{1}p.q^{5} + ^{6}C_{2}p^{2}q^{3}

⇒ 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

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