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1 year ago

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

Mathematics

2 answers:

Lyrx [107]1 year ago

7 0

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

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

<h3><u>Question 1</u></h3>

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,

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

<h3><u>Question 2</u></h3>

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

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

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

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

fgiga [73]1 year ago

4 0

Answer:

See below ~

Step-by-step explanation:

<u>P (6th grader)</u>

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

<u>P (6th grader after)</u>

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

<u>Question 1 : P (Both 6th graders)</u>

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

<u></u>

<u>Question 2 : P' (Both 6th graders)</u>

P' = 1 - P
P' = 1 - 1/14
P' = <u>13/14</u>

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