Question

a basket contains 3 orange 2 yellow 3 green balls if two balls are picked at random, what is the probability that two are yellow?

Answer 1

Answer:

Probability = 1/4

Explanation:

Sample Space: {3o, 2y, 3g}.

n(S) = 8

Event A: Ball Picked is yellow

A ={ 2yellow}

n(A) = 2

p(A) = n(A) / n(S)

= 2 / 8

= 1/4

Therefore, Probability is 1/4

Answer 2

The probability that two selected balls are yellow is $\frac{1}{28}$

Given:

A basket contains 3 orange 2 yellow 3 green balls and two balls are picked at random.

To Find:

The probability that two are yellow.

Solution:

To find the probability that two picked balls are yellow, we will follow the following steps:

According to the question:

Orange ball = 3

Yellow ball = 2

Green ball = 3

Total number of balls = 3 + 2 + 3 = 8 balls

It is not given that ball is picked with or without replacement so, we will take it as without replacement.

Now,

Probability of getting any of the 2 balls from 8 balls =

$8C2$

Combinations are solved as:

$nCr = \frac{n!}{r!(n - r)!}$

Now,

Putting values we get,

$8C2 = \frac{8!}{2!(8 - 2)!} = \frac{8 \times 7 \times 6!}{2 \times 6!} = \frac{8 \times 7}{2} = 4 \times 7 = 28$

Probability of getting 2 yellow balls from 2 yellow balls =

$2C2$

$2C2 = \frac{2!}{2!(2 - 2)!} = \frac{2 \times 1}{2 \times 1 \times 0!} = \frac{2}{2 \times 1} = 1$

Now,

The probability of getting two yellow balls from 8 balls is =

$\frac{probability \: of \: getting \: 2 \: yellow \: balls}{probability \: of \: getting \: any \: of \: the\: two \: balls \: from \: 8 \: balls}$

$\frac{1}{28}$

Henceforth, the probability that two selected balls are yellow is $\frac{1}{28}$

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