Question

Suppose you have a bag with 10 black and 10 red balls, with the balls of each color numbered 1 to 10. Suppose you pick two balls uniformly at random from the bag.

a. Show that the probability that you pick a ball of each color is 10/19.
b. Show that the probability that you pick a ball of each color, with the number on the red ball being one less than the number on the black ball, is 9/190.

Answer 1

Answer:

Step-by-step explanation:

$\text{From the given information:}$

$(a)$

$\text{ P(ball of each color)=P(1st ball is black \& 2nd (red)) + P(1st(red) \& 2nd(black)})$

$\text{ P(ball of each color)} = {P(1st (black) \times P(2nd \ red \ | \ 1st black) }+P(1st \ red) \times P(2nd \ black \ | \ 1st \ red)}$

$\text{ P(ball of each color)} = \Big(\dfrac{10}{20}\Big ) \times \Big(\dfrac{10}{19}\Big ) + \Big(\dfrac{10}{20}\Big ) \times \Big(\dfrac{10}{20}\Big )$

$\text{ P(ball of each color)} = \Big(\dfrac{10}{19}\Big )$

$(b)$

$\text{If the combination of total no. of black \& red balls with no =10*10 =100}$

$Then;$

$\text{the no. of ways red is less than no. of black = 9}$

$Thus:$

$\text{P(ball of each color and number on red ball is 1 less than number of black ball is :)}$$= \text{P(ball of each color) * P(red is 1 less than black [] ball of each color)}$

$=\Big( \dfrac{10}{19}\times \dfrac{9}{100} \Big )$

$= \dfrac{9}{190}$