Question

a box contains 3 white balls and 4 black balls. a ball is drawn at random the color is recorded and then the ball is put back in the box. Then a second ball is drawn at random from the same box. find the probability of the event that at least one of the balls is white

Answer 1

The box has 3 white balls and 4 black balls.

Total number of balls = 3 + 4 = 7

First draw:

The probability of getting a white ball is given by

`\begin{gathered} P(white)=\frac{\text{number of white balls}}{total\text{ number of balls}} \\ P(white)=(3)/(7) \end{gathered}`

Second draw:

Notice that after the first draw the ball is put back in the box.

The probability of getting a white ball is given by

`P(white)=(3)/(7)`

At least one of the balls is white means that one white ball or two white balls.

`P(x\ge1)\; =P(x=1)+P(x=2)_{}`

We have already found the probability of getting one white ball that is P(x=1) = 3/7

The probability of getting two white balls is

`\begin{gathered} P(two\; white)=P(white)* P(white) \\ P(two\; white)=(3)/(7)*(3)/(7)=(9)/(49) \end{gathered}`

Finally, the probability of at least one white ball is

`\begin{gathered} P(x\ge1)\; =P(x=1)+P(x=2)_{} \\ P(x\ge1)\; =(3)/(7)+(9)/(49) \\ P(x\ge1)\; =(30)/(49) \end{gathered}`

Therefore, the probability of the event that at least one of the balls is white is 30/49