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answer
[tex]\frac{33}{248}[/tex] ≈ 0.1331 = 13.31 %
first chocolate
the probability of picking a chocolate the first time is [tex]\frac{solid chocolates}{total chocolates}[/tex]
there are 12 solid chocolates and 32 total chocolates
[tex]\frac{solid chocolates}{total chocolates}[/tex] = [tex]\frac{12}{32}[/tex] = [tex]\frac{3}{8}[/tex]
second chocolate
the probability of picking a chocolate the first time is [tex]\frac{solid chocolates}{total chocolates}[/tex]
since we already picked a solid chocolate the first time, there is one less solid chocolate and one less total chocolate
there are 12 - 1 = 11 solid chocolates and 32 - 1 = 21 total chocolates
[tex]\frac{solid chocolates}{total chocolates}[/tex] = [tex]\frac{11}{31}[/tex]
total probability
to find the probability of both being solid chocolates, multiply their individual probabilities together
[tex]\frac{3}{8} *\frac{11}{31}[/tex]
= [tex]\frac{3 * 11}{8 * 31}[/tex]
= [tex]\frac{33}{248}[/tex]
≈ 0.1331
= 13.31 %
The probability of selecting 2 solid chocolates in a row is 13.31% and this can be determined by using the given data and concept of probability.
Given :
- There are 32 chocolates in a box, all identically shaped.
- There are 9 filled with nuts, 11 with caramel, and 12 are solid chocolate.
The probability to pick one solid chocolate in the first attempt is:
[tex]\rm P = \dfrac{12}{32}[/tex]
[tex]\rm P = \dfrac{3}{8}[/tex]
P = 0.375
The probability to pick another solid chocolate in the second attempt is:
[tex]\rm P' = \dfrac{11}{31}[/tex]
So, the probability of selecting 2 solid chocolates in a row is given by:
[tex]\rm P_T = P + P'[/tex]
[tex]\rm P_T = \dfrac{3}{8}+\dfrac{11}{31}[/tex]
[tex]\rm P_T = \dfrac{33}{248}[/tex]
[tex]\rm P_T = 0.1331[/tex]
[tex]\rm P_T = 13.31[/tex]%
For more information, refer to the link given below:
https://brainly.com/question/21586810
