QUESTION IMAGE
Question
determining probability with candy
name:
use the candy box to solve each problem.
tasty candy
(cherry 1) (cherry 2) (cherry 3)
(lemon) (cherry 4) (lemon)
(lemon) (cherry 5)
(lemon) (cherry 6) (grape 1)
(lemon) (grape 4) (grape 3) (grape 2)
- if you picked 1 piece of candy out of the box which flavor would you have the highest probability of selecting?
- what is the probability of selecting a grape piece?
- if you picked a piece at random would you be more likely to select, a lemon piece or a cherry piece?
- how many total pieces of candy are in the box?
- your friend wants either a cherry piece or a grape piece. if you picked a piece out randomly, which one would you have the highest probability of selecting?
- if you ate 4 lemon pieces, 5 cherry pieces and 3 grape pieces, which flavor would you have the highest probability of selecting next?
- what is the probability of selecting either a cherry piece or a grape piece?
- what is the probability of selecting a lemon piece?
- which flavor has the lowest probability of being selected?
- what is the probability of selecting a cherry piece?
Step1: Count each candy flavor
Cherry: 6, Lemon: 5, Grape: 4
Total candies: $6+5+4=15$
Step2: Solve sub-1 (highest probability)
Cherry count (6) > Lemon (5) > Grape (4)
Step3: Solve sub-2 (grape probability)
Probability = $\frac{\text{Grape count}}{\text{Total}} = \frac{4}{15}$
Step4: Solve sub-3 (lemon vs cherry)
Cherry count (6) > Lemon count (5)
Step5: Solve sub-4 (total candies)
Sum: $6+5+4=15$
Step6: Solve sub-5 (cherry vs grape)
Cherry count (6) > Grape count (4)
Step7: Solve sub-6 (after eating)
Remaining lemon: $5-4=1$, cherry: $6-5=1$, grape: $4-3=1$ → equal
Step8: Solve sub-7 (cherry or grape)
Count: $6+4=10$, Probability = $\frac{10}{15}=\frac{2}{3}$
Step9: Solve sub-8 (lemon probability)
Probability = $\frac{5}{15}=\frac{1}{3}$
Step10: Solve sub-9 (lowest probability)
Grape count (4) is smallest
Step11: Solve sub-10 (cherry probability)
Probability = $\frac{6}{15}=\frac{2}{5}$
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