Question

If 8 balls are distributed at random among three boxes, what is the probability that the first
box would contain exactly 3 balls?​

Answer 1

Subtract 3 from 8 and you will get 5 as probability is always in fraction write 5\8

Answer 2

The Probability That The First Box Would Contain Exactly 3 Balls Is 0.2731

GIVEN

Number of balls = 8

Number of boxes in which balls are distributed= 3

TO FIND

The probability that the first box has 3 balls

SOLUTION.

we can simply solve the above problem as follows-

It is given that the 8 balls are distributed randomly in 3 boxes.

if each box contains 3 balls.

Then,

The total number of ways in which 8 balls can be distributed among 3 boxes = 3 × 3 × 3.........8 times.

We can write it as-

The number of ways = 3⁸

Now, we have to find the probability for 1st box that will contain 3 balls

Number of ways of selecting 3 balls from 8 balls = ⁸C₃

Since we have already selected 3 balls for 1 box

Number of balls left = 8-3 = 5

Number of boxes left = 3-1= 2

Now, the number of ways in which 5 balls can be distributed in 2 boxes = 2×2×2......5 = 2⁵ times

Number of ways of arranging balls such that 1st box has 3 balls and 5 balls get distributed in 2 boxes = ⁸C₃ × 2⁵

we know that,

$probability = \frac{favorable \: outcome}{total \: outcome}$

where,

Favorable outcome = ⁸C₃ × 2⁵

Total outcome = 3⁸

putting the value in the above formula

$= \frac{ ⁸C₃ × 2⁵}{ {3}^{8} }$

$= \frac{56 \times 32}{6561}$

$= \frac{1792}{6561}$

= 0.2731

Hence, The Probability That The First Box Would Contain Exactly 3 Balls Is 0.2731

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