A weighted coin with the probability of showing a head is (2/3) is tossed. If head appears, then a number is selected from the first 20 natural numbers otherwise a number is selected from the first 9 natural numbers. The probability of getting an even number is

Question

A weighted coin with the probability of showing a head is (2/3) is tossed. If head appears, then a number is selected from the first 20 natural numbers otherwise a number is selected from the first 9 natural numbers. The probability of getting an even number is (A) (1/2) (B) (13/27) (C) (10/27) (D) (5/13)

Tardigrade Admin · Accepted Answer

P(H)=(2/3),P(T)=(1/3) A= the selected number is an even number P(A)=P(H)P((A/H))+P(T)⋅ P((A/T)) =(2/3)× (10/20)+(1/3)× (4/9)=(1/3)+(4/27)=(13/27)

Q. A weighted coin with the probability of showing a head is $\frac{2}{3}$ is tossed. If head appears, then a number is selected from the first $20$ natural numbers otherwise a number is selected from the first $9$ natural numbers. The probability of getting an even number is

$\frac{1}{2}$

$\frac{13}{27}$

$\frac{10}{27}$

$\frac{5}{13}$

$P (H) = \frac{2}{3}, P (T) = \frac{1}{3}$
$A =$ the selected number is an even number
$P (A) = P (H) P (\frac{A}{H}) + P (T) \cdot P (\frac{A}{T})$
$= \frac{2}{3} \times \frac{10}{20} + \frac{1}{3} \times \frac{4}{9} = \frac{1}{3} + \frac{4}{27} = \frac{13}{27}$

Q. A weighted coin with the probability of showing a head is 32​ is tossed. If head appears, then a number is selected from the first 20 natural numbers otherwise a number is selected from the first 9 natural numbers. The probability of getting an even number is

21​

2713​

2710​

135​