Question

3. An urn contains 2 black, 2 red, and 4 white marbles. You draw a marble at random. a) What is the sample space? b) What are all events (power Class)? c) What is the probability of each event? 4. A coin is tossed ten times and number of heads is noted. a) What is the probability that the number of heads is four? b) What is the probability of at least seven heads come? 6. X is a Binomial random variable with parameters n and p. Let Z=3X-1. Find the variance of Z. 7. In the last 100 years, 20 big earthquakes were recorded in Turkey. a) Estimate, how many years are there in these 100 years that no earthquake is recorded. b) Estimate, how many years are there in these 100 years that two earthquakes are recorded. PLEASE SOLVE ALL OF THEM I GONNA VOTE UP

Answer 1

Q3) a) Therefore the sample space here is given as:
S = { B1, B2, R1, R2, W1, W2, W3, W4 } that is each of the 8 balls can be taken out here.

b) Drawing one marble from the urn, we can either draw black, red of white marble here. Therefore the power class here is given as:

= { Black, Red, White }
This the power class here,

c) The probability of each event is computed here as:
P(Black) = 2/8 = 0.25,
P(Red) = 2/8 = 0.25,
P(White) = 4/8 = 0.5

Q4) a) The probability that the number of heads in 10 tosses is 4 is computed here as:

$= \binom{10}{4}0.5^{10} = 0.2051$

Therefore 0.2051 is the required probability here.

b) The probability of at least 7 heads is computed here as:

$P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X =10)$

$P(X \geq 7) = 0.5^{10}(\binom{10}{7} + \binom{10}{8} + \binom{10}{9} + 1) = 0.1719$

Therefore 0.1719 is the required probability here.

Q6) We are given the distribution here as:

$X \sim Bin(n, p)$

Therefore, Var(X) = np(1-p)

Z = 3X - 1

Therefore,
Var(Z) = Var(3X -1) = 3²Var(X) = 9*np(1-p)

Therefore 9np(1-p) is the required variance here.

Q7) The mean number of earthquakes in 100 years is given as 20. Therefore the mean number of earthquakes in each year here is modelled as:

N~ Poisson (20/100)

$N \sim Poisson(0.2)$

a) The estimated number of years with no earthquakes in 100 years is computed here as:

= e^-0.2 * 100

= 81.87

Therefore 81.87 is the required expected number of years here.

b) For 2 earthquakes in a year, the probability is computed here as:

$= \frac{0.2^2}{2}e^{-0.2} = 0.01637$

Therefore expected number of years:
= 100*0.01637 = 1.637

Therefore 1.637 is the expected number of years here.