Question

There are 6 sixth graders , 7 seventh graders , and 8 eight graders entered in a contest.

Answer 1

Answer:

$\sf 1) \quad \dfrac{1}{14}$

$\sf 2) \quad \dfrac{13}{14}$

Step-by-step explanation:

Given:

• 6 sixth graders
• 7 seventh graders
• 8 eight graders

Total = 6 + 7 + 8 = 21

$\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}$

Question 1

The probability of the 1st pick being a 6th grader:

$\implies \sf P(6th\:grader)=\dfrac{6}{21}=\dfrac{2}{7}$

Now there will be 5 sixth graders left and a total of 20 left.

So, the probability of the 2nd pick being a 6th grader:

$\implies \sf P(6th\:grader)=\dfrac{5}{20}=\dfrac{1}{4}$

Therefore,

$\implies \textsf{P(6th grader) and P(6th grader)}= \sf \dfrac{2}{7} \times \dfrac{1}{4}=\dfrac{2}{28}=\dfrac{1}{14}$

Question 2

Law of Total Probability states that the sum of probabilities is 1

$\implies \textsf{P(two 6th graders) + P(not two 6th graders)}=1$

$\implies \sf \dfrac{1}{14}+\textsf{P(not two 6th graders)}=1$

$\implies \sf \textsf{P(not two 6th graders)}=1-\dfrac{1}{14}=\dfrac{13}{14}$

Answer 2

Answer:

See below ~

Step-by-step explanation:

P (6th grader)

• No. of 6th graders / Total students
• 6 / 6 + 7 + 8
• 6/21
• 2/7

P (6th grader after)

• No. of 6th graders - 1 / Total students - 1
• 6 - 1 / 21 - 1
• 5/20
• 1/4

Question 1 : P (Both 6th graders)

• P = P (6th grader) × P (6th grader after)
• P = 2/7 x 1/4 = 2/28 = 1/14

Question 2 : P' (Both 6th graders)

• P' = 1 - P
• P' = 1 - 1/14
• P' = 13/14