Let X denote the number of white balls drawn from the urn.
Since there are 4 white balls, X can take values 0,1,2,3,4.
P(X = 0)=p(gettmg no white balls)\(=\frac { { 6C }_{ 4 } }{ 10{ C }_{ 4 } } \)
\(\frac{1}{14}\)
P(X = 1) = P(getting one white ball and 3 red balls) \(\frac { { 4C }_{ 4 }\times { 6C }_{ 3 } }{ { 10 }C_{ 4 } } =\frac { 8 }{ 21 } \)
P(X = 2) = P(getting two white balls and 2 red balls) \(\frac { { 4C }_{ 2 }\times { 6C }_{ 2 } }{ { 10 }C_{ 4 } } =\frac { 8 }{ 21 } \)
P(X = 3) = P(getting 3 white balls and 1 red ball) \(\frac { { 4C }_{ 3 }\times { 6C }l }{ { 10 }C_{ 4 } } =\frac { 4 }{ 35 } \)
P(X = 4) = P(getting 4 white balls) =\(\frac { 4{ C }_{ 4 } }{ 10{ C }_{ 4 } } \)
=\(\frac{1}{210}\)
Thus the probability distribution of X is

Let X denote the number of aces in a pack of 52 playing cards.
Since there are 4 aces in a pack, and we are drawing 3 ace cards, X can take values 0, 1, 2, 3.
P(X = 0) = P(getting no ace card) \(\frac { 48{ C }_{ 3 } }{ 52{ C }_{ 3 } } \)
\(=\frac { 4324 }{ 5525 } \)
P(X=1) = P(getting one ace card and 2 other cards)
\(\\ =\frac { 4{ C }_{ 1 }\times 48{ C }_{ 2 } }{ 52{ C }_{ 3 } } =\frac { 1128 }{ 5525 } \)
P(X=2) = P(getting 2ace card and one other card)
\(=\frac { { 4C }_{ 2 }\times { 48C }_{ 1 } }{ 52{ C }_{ 3 } } =\frac { 72 }{ 5525 }\)
P(X=3) = P(getting 3 ace card) = \(\frac { { 4C }_{ 3 } }{ { 52C }_{ 3 } } \)
\(=\frac { 1 }{ 5525 } \)
Hence, the probability distribution function is

We have P(X=r)∝α^r
⇒P(X = r) = λα^r,r = 0,1,2,....
Since sum of all the probabilities in a probability distribution is 1.
P(X = 0)+P(X = 1)+P(P(X = 2)+... = 1
⇒ λα0+λα1+λα2+...= 1
⇒λ(1+α+α2+.....) = 1
⇒\(\lambda(\frac{1}{1-\alpha})\) = 1
⇒ λ = 1 - α
∴ P(X = r) = (1-α)α^r, r = 0,1,2,...
Hence P(X=0) = (1-α)α⁰ = (1-α)(1) = 1-α.

The probability distribution of X is

Σp_i = 1 ⇒ 0.1 + k + 2k + 2k + k = 1
⇒ 0.1 + 6k = 1 ⇒ 6k = 1-0.1 = 0.9
⇒ k = \(\frac{0.9}{6}\) = 0.15
And the probability that you study atleast 2 hours is P(X≥2)=P(X=2)+P(X=3)+P(X=4)
= 2k + 2k + k = 5k
= 5(0.15) = 0.75

Since the random variable X is the probability distribution function, Σp_i = 1
∴ a + 2a + 3a + 4a + 5a + 6a = 1
21a = 1 ⇒ a = \(\frac{1}{21}\)
Now, F(4) = P(X ≤ 4)
= P(X = 0) + P(X = 1) + P(X = 2)P(X = 3) + P(X = 4)
= a + 2a + 3a + 4a + 5a = 15a
= 15\((\frac{1}{21})=\frac{5}{7}\)
∴ F(4) = \(\frac{5}{7}\)