Math, asked by bishtleela12gmailcom, 6 hours ago

A teacher conducted a fun activity in the class. She put cards numbered from 9 to 90 in a box. Then she called students one by one,from the teams formed by her.The child speaks out any property related to numbers.If he gets a number satisfied that property,the team scored marks otherwise not.

(i) Vanshika speaks out 'divisible by 6'.The probability that her team get marks is
(a)1/6 (b)7/41 (c)7/45 (d)5/27

(ii) Sam speaks out 'a perfect square number'.The Probability of his team getting marks is
(a)7/82 (b)8/82 (c)10/82 (d)7/90

(iii) Preeti says 'a prime number'.The probably of her team getting marks is
(a)19/82 (b)11/41 (c)10/41 (d)21/82

(iv) Karan says 'an odd number'.The probability of getting score in this cases is
(a)20/41 (b)1/2 (c)1/4 (d)1/3

(v) Which of the following property will have maximum chances of getting marks?
(a)an even prime number (b)an even number
(c)a one digit number (d)a two-digit number​

Answers

Answered by satyamvishwakarma463
1

Answer:

total outcome=90

favourable outcome=let x

Answered by Hansika4871
2

Given:  

A set of cards numbered from 9 to 90 and are put into a box.

Every card is picked and a clue is given about the picked card.

To Find:

1. Probability of student's team getting correct if the drawn card is divisible by 6.

2. Probability of student's team getting correct if the drawn card is a perfect square number.

3. Probability of student's team getting correct if the drawn card is a prime number.

4. Probability of student's team getting correct if the drawn card is an odd number.

5. Probability of higher chance of mentioned categories.

Solutions:

1. Probability of an event is defined as the favorable outcomes divided by the total number of outcomes.

P(Event happening) = (Number of events favoring)/(Total number of events)

Question (i)- OPTION B:

  • Probability of the team answer for getting marks if the picked number is divisible by 6:
  • Favorable outcomes are : ( 12,18,24,30,36,42,48,54,60,66,72,78,84,90)
  • Total number of Outcomes are 82. ( 9 to 90 both inclusive)

=>Probability of getting right :

(Total numbers which are divisible by 6)/(total number of outcomes)

=> P(getting right) =  14/82 ,

=> P(getting right) = 7/41.

Therefore, the probability of the student's team getting the correct answer is 7/41.

Question (ii) - Option A:

  • Probability of the team getting correct if the picked card is a perfect square number:
  • Total favorable outcomes:(9,16,25,36,49,64,81) (7 favorable outcomes)
  • Total number of outcomes is 82.

=>Probability of the student's team to get right:

(Total numbers which are perfect squares)/(Total number of outcomes)

=> Probability of student's team to get right = 7/82.

Therefore, the probability that the student's team will get marks for this case is 7/82.

Question (iii) - Option c:

  • Probability of the team getting marks if the picked card is a prime number:
  • Total favorable outcomes: (11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89)
  • Total number of outcomes is 82.

=> Probability for getting marks: ( Total favorable cases/ Total number of cases)

=> Probability of getting marks = 20/82= 10/41.

Therefore, the probability for the student's team to get a score, in this case, is 10/41.

Question (iv) - Option b:

  • Probability for the student's team to get a score if the picked card is an odd number:
  • Total favorable outcomes are: (odd numbers between 9 and 90 i.e 50 outcomes).
  • Total possible outcomes = 82.

=> Probability of getting right = (Favorable outcomes / Total number of Outcomes)

=> Probability of getting right = 41/82 = 1/2.

Therefore, The probability for the student's team to get score in this case is 1/2.

Question (v) - Option d:

a. There are no even prime numbers after 9 and below 90, Therefore the favorable outcomes are 0. Hence the probability, in this case, is 0.  

b. There are a total of 41 even numbers between 9 and 90, Therefore favorable outcomes are 41.

  • P(an even number) = 41/82 = 1/2.

c.  There is only one single-digit number from 9 to 90, Therefore favorable outcomes are 1.

  • P(single digit number) = 1/82.

d. There are 81 two-digit numbers from 9 and 90, Therefore favorable outcomes are 81.

  • P(getting a two-digit number) = 81/82.

Therefore, Option d has the highest Probability, Hence it has the highest chance of getting.

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