Question

A box contains 4 chocobars and 4 ice-creams. Tom eats 3 of them one after another. What is the probability of sequentially choosing 2 chocobars and 1 ice-cream? Correct to 2 decimal places.

Answer 1

Given:

The number of choco bars = 4.

The number of ice creams = 4.

Aim:

We need to find the probability of sequentially choosing 2 choco bars and 1 ice cream.

Explanation:

The total items in the box =4+4 =8.

$n(S)=8$

Let A be the event that chooses 2 choco bars.

There are 4 choco bars in the box.

$n(A)=4$

The probability of choosing 2 choco bars.

$P(A)=\frac{n(A)}{n(S)}$

Substitute known values,

$P(A)=\frac{4}{8}=\frac{1}{2}$

After taking 2 chocobars, the number of items in box = 8-2 =6.

$n(S_1)=6$

Let B be the event that takes one ice cream from the box.

There are 4 ice creams in the box.

$n(B)=4$

The probability of choosing one ice cream.

$P(B)=\frac{n(B)}{n(S_1)}$

Substitute known values.

$P(B)=\frac{4}{6}=\frac{2}{3}$

The probability of sequentially choosing 2 choco bars and 1 ice-cream

$=P(A)\times P(B)$$=\frac{1}{2}\times\frac{2}{3}$$=0.33$

Final answer:

The probability of sequentially choosing 2 choco bars and 1 ice cream is 0.33.