Question

Balls numbered 1 through 10 are placed in a bag. Three balls are drawn out of the bag with replacement. What is the probability that all the balls have even numbers on them?

Answer 1

Step-by-step explanation:

There are 20 balls in a bag.

Odd numbers between 1-20 are {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}=10

The probability of getting 3 odd numbered balls without replacement:

$= > \frac{10}{20} \times \frac{9}{19} \times \frac{8}{18} \\ \\ = > \frac{720}{6840} \\ \\ = > 0.105$

Answer 2

$\huge{\mathfrak{\underline{\red{Answer}}}}$

━━━━━━━━━━━━━━━

$\mathbb{\boxed{\pink{GIVEN}}}$

• Balls are numbered 1 to 10
• One ball is drwan out at random.

━━━━━━━━━━━━━━━━

Then the numbers marked on the ball can be either of

• 1
• $\underline{\boxed{\green{2}}}$
• 3
• $\underline{\boxed{\green{4}}}$
• 5
• $\underline{\boxed{\green{6}}}$
• 7
• $\underline{\boxed{\green{8}}}$
• 9
• $\underline{\boxed{\green{10}}}$

here only the boxed numbers are even.

The balls can be

1,2,3

2,3,4

3,4,5

4,5,6

5,6,7

6,7,8

7,8,9

8,9,10

1,3,4

1,4,5....

in this way there can be 120 ways

Thus the number of balls having even number on them can be

2,4,6

4,6,8

6,8,10

4,6,10

4,2,8

6,2,10......

in this way 40 types of 3 balls with even number can be picked up

Total number of outcome= 1000

Thus the probability is:

$\frac{no \: of \: balls \: with \: even \: number}{total \: number \: of \: balls} \\ \implies \: \frac{40}{120} \\ \implies \frac{1}{3}$

$\huge{\boxed{\red{Probability=\:1/3}}}$