Question

There are 36 chocolates in a box, all identically shaped. there 10 are filled with nuts, 12 with caramel, and 14 are solid chocolate. you randomly select one piece, eat it, and then select a second piece. find the probability of selecting a solid chocolate candy followed by a nut candy.

Answer 1

Total chocolates = 36
Total with nuts = 10
Total with caramel = 12
Total with Solid chocolate = 14

P(Solid Chocolate) = 14/46 = 7/18

Answer: 7/18

Answer 2

Answer: $\dfrac{1}{9}$

Step-by-step explanation:

Given : The total number of chocolates in the box = 36

The number of chocolates filled with solid chocolate  =14

The number of chocolates filled with nuts = 10

Now, the probability of selecting a solid chocolate candy is given by :-

$=\dfrac{14}{36}$

Now the chocolates left in box = 36-1=35

The probability of selecting next chocolate a nut candy will be :-

$=\dfrac{10}{35}$

Now, the probability of selecting a solid chocolate candy followed by a nut candy will be:_

$=\dfrac{14}{36}\times\dfrac{10}{35}=\dfrac{1}{9}$