contestada

there are 6 sixth graders , 7 seventh graders , and 8 eight graders entered in a contest.
Each of their names is written on a slip of paper and the all the slips are placed in a basket. prizes are given to the 2 students whose names are randomly pulled from the basket.

1. what is the probability that both winners will be 6th graders


2. what is the probability that both winners will NOT be 6th graders

Respuesta :

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
semsee45

Answer:

[tex]\sf 1) \quad \dfrac{1}{14}[/tex]

[tex]\sf 2) \quad \dfrac{13}{14}[/tex]

Step-by-step explanation:

Given:

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

Total = 6 + 7 + 8 = 21

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

Question 1

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

[tex]\implies \sf P(6th\:grader)=\dfrac{6}{21}=\dfrac{2}{7}[/tex]

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:

[tex]\implies \sf P(6th\:grader)=\dfrac{5}{20}=\dfrac{1}{4}[/tex]

Therefore,

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

Question 2

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

[tex]\implies \textsf{P(two 6th graders) + P(not two 6th graders)}=1[/tex]

[tex]\implies \sf \dfrac{1}{14}+\textsf{P(not two 6th graders)}=1[/tex]

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

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