Bernoulli trials is also called a Dichotomous experiment and is repeated n times. Bernoulli trials is also called as a ?yes? or ?no? questions because it has only two outcomes, those are ?success? and ?failure?.

Probability of getting 3 in first throw = \(\frac {1}{6}\)
Probability of getting of 3 in second throw = \(\frac {1}{6}\)
Total probability = \(\frac {1}{6} \times \frac {1}{6} = \frac {1}{36}\)

We know that P(S|F) = P(S∩F) / P(F). (By formula for conditional probability)
Which is equivalent to P(F|F) = P(F) / P(F) = 1, hence value of P(F|F) = 1.

Total number of balls = 4 + 2 + 7 = 13
The probability of first ball to be blue = \(\frac {7}{13}\)
The probability of drawing red ball as second ball = \(\frac {4}{13}\)
Total probability = \(\frac {7}{13} \times \frac {4}{13} = \frac {28}{169}\)

Possible outcomes when a dice is thrown = {1,2,3,4,5,6}\(\frac {number \, of \, odd \, numbers}{number \, of \, possible \, outcomes} = \frac {3}{6}=\frac {1}{2}\)

Total number of balls = 9
The probability of first ball to be red = \(\frac {3}{9}\)
The probability of second ball to be green with replacement = \(\frac {4}{9}\)

Let E1 = event of choosing bag 1, E2 = event of choosing bag 2.
Let A be event of drawing a red ball.
P(E1) = P(E2) = 1/2.
Also, P(A|E1) = P(drawing a red ball from Bag 1) = 3/8.
And P(A|E2) = P(drawing a red ball from Bag 2) = 4/10.
The probability of drawing a ball from bag 2, being given that it is red is P(E2|A).
By using Bayes? orem,
P(E2|A) = P(E2)P(A|E2)/( P(E1)P(A?E1)+P(E2)P(A|E2))
= (1/2 × 4/10) / ((1/2 × 3/8)) + (1/2 × 4/10)) = 16/31.

Formula for binomial distribution is P [X = x] = ⁿC_x p^x q^n-x
Where n is number of trials, x is number of successes and (n-x) is failures.

Poisson distribution shows number of times an event is likely to occur within a specified time. The Poisson distribution probability formula is P(x; μ) = (e^-μ) (μ^x) / x!

Since 6 balls are green, probability of first ball to be drawn is green = \(\frac {6}{9}\)
The probability of first ball to be green, 8 balls are left and 5 out of m are green = \(\frac {5}{8}\)
If first two balls are green, n probability of third ball to be green = \(\frac {4}{7}\)
According to multiplication orem, \(\frac {6}{9} . \frac {5}{8} . \frac {4}{7} = \frac {5}{21}\)