Two of the desserts are sugar/free. Three of the drinks are sugar-free. A meal deal combination is selected at random. Work out the probability that both the desserts and the drink are sugar free. Give your answer as a fraction

Respuesta :

Answer: 8/15

Step-by-step explanation:

You need part a of the question to answer this question correctly. In part a of the question it is said there is 6 sandwiches, 3 desserts and 5 drinks.

Two of the desserts are sugar free (2/3).

Four of the drinks are sugar free (4/5)

Times these fractions together, 2/3x4/5= 8/15.

Hope this helps :)

fichoh

The probability of selecting a sugar free dessert and drink is 2/5

Number of sandwiches = 6

Number of dessert = 3

Number of drinks = 5

Sugar free drinks = 3

Sugar free dessert = 2

Recall :

Probability = required outcome / Total possible outcomes

P(both dessert and drink are sugar free) :

(sugar free dessert) × p(sugar free drink)

(2 /3) × (3 / 5) = 6/15 = 2 / 5

Hence, the probability of selecting a sugar free dessert and drink is 2/5

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