Answer to Question #241066 in Statistics and Probability for Abir

Question

1. 20% of people are vegetarians. 70% of vegetarians like broccoli. 20% of vegetarians like
carrot. 10% of vegetarians like cucumber. 40% of meat eaters like broccoli. 60% of meat
eaters like carrot. What is the probability that a man who likes carrot is a meat eater?
2. Tossing 2 coins together, if at least one head comes roll a die else toss again. Find the
probability distribution for the random variable X which represents the number of tail as
output.
3. What is the coefficient of x⁹y⁶ in (x-y)
4. Explain Euclid’s Algorithm with an example?

Answer 1

Part 1

The probability that a man who likes carrots is a meat eater

$=(\frac{80}{100} \times \frac{60}{100}) \times (\frac{20}{100}\times \frac{80}{100})\\ =\left(\frac{4}{5}\times \frac{3}{5}\right)\times \left(\frac{20}{100}\times \frac{4}{5}\right)\\ =\frac{12}{25}\times \frac{4}{25}\\ =\frac{48}{625}$

Part 2

Back-up Theory

If A and B are independent, P(A and B) = P(A ∩ B) = P(A) x P(B) ..……….....................…(1)

Now to work out the solution,

Given,

X = number of tosses until one of the two coins turns up a head and the other a tail.

P(the first coin turns up head) = 0.74 ....................................................................................... (2)

P(the second coin turns up head) = 0.12 ................................................................................. (3)

The pay-off at the end of the game: G(X) = X² + 2 ....................................................................(4)

All tosses are independent. ................................................................................................ (5)

We want:

$P[G(X) > 6]\\ = P[X2 + 2 > 6] [vide (4)]\\ = P[X2 > 4]\\ = P(X > 2) [note X2 > 4 => X > 2 or X < - 2; but then X cannot be negative]\\ = 1 – P(X ≤ 2)\\ = 1 – P(X = 2)$

[‘one of the two coins turns up ahead and the other a tail’ => minimum value of X is 2 ]

= 1 – [P{(the first coin turns up head and the second coin turns up tail)}

        + P{(the first coin turns up tail and the second coin turns up head)}]

= 1 – {(0.74 x 0.88) + ((0.26 x 0.12))} [vide (5), (1), (2) and (3)]

= 1 – 0.6824

= 0.3176

Part 3

$\frac{\partial \:}{\partial \:x}\left(x^9y^6\right)=9x^8y^6\\ \frac{\partial \:}{\partial \:y}\left(x^9y^6\right)=6x^9y^5$

Part 4

$=> GCD(493, 899 - 493) = GCD(493, 406)\\ => GCD(493 - 406, 406) = GCD(87, 406)\\ => GCD(87, 406 - 87) = GCD(87, 319)\\ => GCD(87, 319 - 87) = GCD(87, 232)\\ => GCD(87, 232 - 87) = GCD(87, 145)\\ => GCD(87, 145 - 87) = GCD(87, 58)\\ => GCD(87 - 58, 58) = GCD(29, 58)\\ => GCD(29, 58 - 29) = GCD(29, 29)\\ => GCD(29, 29 - 29) = GCD(29, 0)\\ GCD(29, 0) is 29$