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

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) * P(2nd \ red \ +P(1st \ red) * P(2nd \ black \ | \ 1st \ red)}`

`\text{ P(ball of each color)} = \Big((10)/(20)\Big ) * \Big((10)/(19)\Big ) + \Big((10)/(20)\Big ) * \Big((10)/(20)\Big )`

`\text{ P(ball of each color)} = \Big((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( (10)/(19)* (9)/(100) \Big )`

`= (9)/(190)`