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A resturant serves a bowl of candies to their costumers.the bowl of candies gabriel recieves has 10 chocholate candies,8 coffee candies,and 12 caramel candies. After gabriel chooses a candy,he eats it. Find the probability of getting candies with the indicated flavors. A. P(chocolate or coffee) b. P(caramel or not coffee) c. P(coffee or caramel) d. P(chocolate or not caramel)

Sagot :

CebuMath

Answer:

A. 60%

B. 73%

C. 67%

D. 60%

Step-by-step explanation:

Probability is a topic in mathematics that deals with the possibility of an event to occur. The value of the probability is between 0 and 1 inclusive. It should not go beyond 1 and should go lesser than 0.

Notable Problems in Probability

1. The "or" problem.

2. The "and" problem.

In probability, it is very important that you know what operation to use for a specific problem. If the probability problem deals with an "or" problem, then use addition. If the probability problem deals with an "and" problem, then use multiplication.

Remember that the probability of an event to occur is given by [tex]P(E)=\frac{\text{number of considered outcomes}}{\text{total possible outcomes}}[/tex]. The total candies that are in the bow is 30.

Problem A

The probability of having a chocolate candy is [tex]\frac{10}{30}[/tex].

The probability of having a coffee candy is [tex]\frac{8}{30}[/tex].

Thus, the probability of getting a chocolate candy or coffee candy is given by

                                    [tex]\begin{aligned}\frac{10}{30}+\frac{8}{30}&=\frac{18}{30}\\&=\frac{3}{5}\:\text{or}\:0.6\:\text{or}\: {60\%}\end{aligned}[/tex].

Problem B

The probability of having a caramel candy is [tex]\frac{12}{30}[/tex].

The probability of having a noncoffee candy is [tex]\frac{10}{30}[/tex].

Thus, the probability of getting a caramel candy or noncoffee candy is given by

                                   [tex]\begin{aligned}\frac{12}{30}+\frac{10}{30}&=\frac{22}{30}\\&=\frac{11}{15}\:\text{or}\:{.73}\:\text{or}\:{73\%}\end{aligned}[/tex].

Problem C

The probability of having a coffee candy is [tex]\frac{8}{30}[/tex].

The probability of having a caramel candy is [tex]\frac{12}{30}[/tex].

Thus, the probability of getting a coffee candy or caramel candy is given by

                                   [tex]\begin{aligned}\frac{8}{30}+\frac{12}{30}&=\frac{20}{30}\\&=\frac{2}{3}\:\text{or}\:{.67}\:\text{or}\:{67\%}\end{aligned}[/tex].

Problem D

The probability of having a chocolate candy is [tex]\frac{10}{30}[/tex].

The probability of having a noncaramel candy is [tex]\frac{8}{30}[/tex].

Thus, the probability of getting a chocolate candy or noncaramel candy is given by

                                  [tex]\begin{aligned}\frac{10}{30}+\frac{8}{30}&=\frac{18}{30}\\&=\frac{3}{5}\:\text{or}\:{.6}\:\text{or}\:{60\%}\end{aligned}[/tex].

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To learn more about probabilities, go to

https://brainly.ph/question/1568178

https://brainly.ph/question/2110202

https://brainly.ph/question/2058361

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