Question

There is a bag with only milk and dark chocolates.
The probability of randomly choosing a dark chocolate is 7/9There are 21 dark chocolates in the bag and each is equally likely to be chosen.
Work out how many milk chocolates there must be.

Answer 1

Probability of choosing dark chocolate = 7/9

Number of dark chocolate = 21

7/9 = 21

1/9 = 21/7

1/9 = 3

Total chocolates in the bag = 3×9 = 27

Number of milk chocolates = 27 - 21

= 6

Hope this will help.....

Answer 2

Answer:

The number of milk chocolates in the bag = 6

Step-by-step explanation:

Given,

The bag contains only milk chocolates and dark chocolates

Probability of randomly choosing dark chocolate =$\frac{7}{9}$

No of dark chocolates in the bag = 21

To find,

No of milk chocolates in the bag

Since, Bag contains only milk chocolates and dark chocolates,

Total number of chocolates in the bag =  no of dark chocolates+ no of milk chocolates

We know,

Probability of occurring an event = $\frac{no\ of\ favorable\ items}{total\ number\ of\ items}$

Hence the probability of randomly choosing dark chocolate = $\frac{no\ of\ dark\ chocolates}{total\ number\ chocolates}$

Since the probability of randomly choosing dark chocolate = $\frac{7}{9}$

and no of dark chocolate is 21

We have

$\frac{7}{9}$ = $\frac{21}{total\ number\ of\ chocolates\ in\ the\ bag}$

Total no of chocolates in the bag = $21X \frac{9}{7}$ = 27

Hence the number of milk chocolates in the bag = total no. of chocolates - no of dark chocolates

=27 -21 = 6

The number of milk chocolates in the bag = 6