Q. 225.0( 1 Vote )

# A random variable

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

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