Probability: Play with numbers (part 1: The 10 easiest answered-questions)

Q1. If A and B are two positive integers then what is the probability that A+B is an in even number?
a. ¼ b. ½ c. ⅓ d. ¾
correct: b, answer:
The sum of two odd is even and sum of two even is also even. Sum of an odd and an even is odd and sum of an even and an odd is also even. So, the sample space contains 4 elements.
So probability of sum of two positive integers is even =2/4=½.

Q2. Two numbers are chosen at random. What is the probability that either both of them are even or both of them are odd?
a. ¼ b. ½ c. ¾ d. None of the above
correct: b, answer:
The two numbers can be
Odd odd
Odd even
Even odd
Even even
Number of favourable outcomes =2. The probability equals =½.

Q3. From a box containing slips numbered 1,2,3,…,5 one slip is picked up. Find the probability in each case:
Events Happening:
(i). The number on the stip is a prime number.
answer:

Sample space ={1,2,3,4,5}
=n(S)=5

Let event A denote that the number on the slip is prime number

(ii). The number on the slip is a multiple of 3
answer:
Let event B denote that the no. on the slip is a multiple of 3

Q4. Box A contains 3 cards numbered 1, 2 and 3.
Box B contains 2 cards numbered 1 and 2.
One card is removed at random from each box.
Find the probability that:
a) the sum of the numbers is 4.
answer:
S={(1,1);(1,2);(2,1);(2,2);(3,1);(3,2)}
P=2/6=⅓

b) the sum of the two numbers is a prime number.
answer:
P=4/6=⅔

c) the product of the two numbers is at least 3.
answer:
P=3/6=½

d) the sum is equal to the product.
answer:
P=⅙

Q5. Nine cards (identical in all respects) are numbered 2 to 10. A card is selected from them at random. Find the probability that the card selected will be:
(i) an even number multiple
(ii) a multiple of 3
(iii) an even number and a multiple of 3 (iv) an even number or a multiple of 3.
answer:
There are 9 cards from which one card is drawn.
Total number of elementary events =n(S)=9
(i) From numbers 2 to 10, there are 5 even numbers i.e. 2, 4, 6, 8, 10.
Number of favorable events =n(E)=5
Probability of selecting a card with an even number


(ii) From numbers 2 to 10. there are 3 numbers which are multiples of 3 i.e. 3, 6, 9.
Number of favorable events =n(E)=3
Probability of selecting a card with a multiple of 3

(iii) From numbers 2 to 10. there is one number which is an even number as well as multiple of 3 i.e. 6
Number of favorable events =n(E)=1
Probability of selecting a card with a number which is an even number as well as multiple of 3

(iv) From numbers 2 to 10, there are 7 numbers which are even numbers or a multiple of 3 i.e. 2, 3, 4, 6, 8, 9, 10.
Number of favorable events =n(E)=7
Probability of selecting a card with a number which is an even number or a multiple of 3

Q6. Larry chooses a date at random in January for a party.

Find the probability that he chooses:
(i) a Wednesday.
(ii) a Friday.
(iii) a Tuesday or a Saturday.
Number of possible outcomes =number of days in the month =31
n(S)=31

(i) E= event of selection of a Wednesday ={1,8,15,22,29}
n(E)=5

Probability of selection of a Wednesday

(ii) E= event of selection of a Friday ={3,10,17,24,31}
n(E)=5

Probability of selection of a Friday

(iii) E= event of selection of a Tuesday or a Saturday ={4,7, 11,14,18,21,25,28)
n(E)=8

Probability of selection of a Tuesday or a Saturday

Q7. A game consists of spinning arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8.
spinning arrow a

If the outcomes are equally likely. find the probability that the pointer will point at:
(i) 6 (iv) a number greater than 8
(ii) an even number (v) a number less than or equal to 9
MO a prime number (vi) a number between 3 and 11
answer:
Total number of possible outcomes =12
(i) Number of favorable outcomes for 6=1
P(the pointer will point at 6) =1/12
(ii) Favorable outcomes for an even number are 2, 4, 6, 8, 10, 12. Number of favorable outcomes =6
P(the pointer will be at an even number) =6/12=½
(iii) Favorable outcomes for a prime number are 2, 3, 5, 7, 11. Number of favorable outcomes =5
P(the pointer will be at a prime number) =5/12
(iv) Favorable outcomes for a number greater than 8 are 9, 10, 11, 12. Number of favorable outcomes =4
P(the pointer will be at a number greater than 8) =4/12=⅓
(v) Favorable outcomes for a number less than or equal to 9 are 1, 2, 3, 4, 5, 6, 7, 8, 9. Number of favorable outcomes =9
P(the pointer will be at a number less than or equal to 9) =9/12=¾
(vi) Favorable outcomes for a number between 3 and 11 are 4, 5, 6, 7, 8, 9, 10. Number of favorable outcomes =7
P(the pointer will be at a number between 3 and 11) =7/12

Q8. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, 1, 2, 3, …, 12 as shown in Fig. below. What is the probability that it will point to:

spinning arrow b

(i) 10? (ii) an odd number? (iii) a number which is multiple of 3? (iv) an even number?
Sol:
Total number of possible outcomes n(S)=1, {1, 2, 3,…., 12}
(i) Let E→ event of pointing 10
Number of favorable outcomes n(E)=1, it is only {10}

(ii) Let E→ event of pointing at an odd number
Number of favorable outcomes n(E)=6,{1,3,5,7,9,11}

(iii) Let E→ event of pointing at a number multiple of 3
Number of favorable outcomes n(E)=4,{3,6,9,12}

(iv) Let E→ event of pointing at an even number
Number of favorable outcomes n(E)=6,{2,4,6,8,10,12}

Q9. A learner finds a textbook that has 100 pages. He then selects one page from the textbook. What is the probability that the page has an odd page number?
Write your answer as a decimal (correct to 2 decimal places).
answer:
n(E)= number of outcomes in the event set =50
n(S)= number of possible outcomes in the sample space =100
Finally, we calculate the probability:

Therefore, the probability that the page has an odd page number =0.50.

Q10. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears
(i) a two-digit number (ii) a perfect square number
(iii) a number divisible by 5.
Answer 18:
Total number of discs =90
(i) Total number of two-digit numbers between 1 and 90 is equal to 81.
P(getting a two-digit number)=81/90=9/10=0.9
(ii) Perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81. Therefore, total number of perfect squares between 1 and 90 is 9.
P(getting a perfect square)=9/90=1/10
(iii) Numbers that are between 1 and 90 and divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, and 90. Therefore, total numbers divisible by 5 is equal to 18.
Probability of getting a number divisible by 5 is equal to =18/90=⅕.

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