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Q1 (4)   Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. 13 cm, 12 cm, 5 cm

 

In case of a right triangle, the length of its hypotenuse is highest.  hypotenuse be h. Taking, 5cm, 12 cm By Pythagoras theorem, = given third side. Hence, it is a right triangle with h=13 cm.

10  The number lock of a suitcase has \small 4 wheels, each labelled with ten digits i.e., from \small 0 to \small 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Given, Each wheel can be labelled with 10 digits.  Number of ways of selecting 4 different digits out of the 10 digits =  These 4 digits can arranged among themselves is  ways.  Number of four digit numbers without repetitions = Number of combination that can open the suitcase = 1  Required probability of getting the right sequence to open the suitcase =  

9.(ii) If \small 4-digit numbers greater than  \small 5,000  are randomly formed from the digits \small 0,1,3,5 and \small 7, what is the probability of forming a number divisible by \small 5 when, (ii) the repetition of digits is not allowed?

(ii) Since 4-digit numbers greater than 5000 are to be formed, The  place digit can be filled up by either 7 or 5 in  ways Since repetition is not allowed,  The remaining 3 places can be filled by remaining 4 digits in   ways.   Total number of 4-digit numbers greater than 5000 = We know, a number is divisible by 5 if unit’s place digit is either 0 or 5. Case 1. When digit at  place is 5, the...

9.(i)  If \small 4-digit numbers greater than \small 5000 are randomly formed from the digits \small 0,1,3.5 and \small 7, what is the probability of forming a number divisible by \small 5 when, (i) the digits are repeated? 

(i) Since 4-digit numbers greater than 5000 are to be formed, The  place digit can be filled up by either 7 or 5 in  ways Since repetition is allowed,  Each of the remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 in  ways.   Total number of 4-digit numbers greater than 5000 =  (5000 cannot be counted, hence one less) We know, a number is divisible by 5 if unit’s place...

8. From the employees of a company, \small 5 persons are selected to represent them in the  managing committee of the company. Particulars of five persons are as follows:     

S. No. Name Sex Age in years
1. Harish M 30
2. Rohan M 33
3. Sheetal F 46
4. Alis F 28
5. Salim M 41

A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over \small 35 years?
 

 

Given, Total number of persons = 5 No. of male spokesperson  = 3 No. of spokesperson who is over 35 years of age = 2 Let E be the event that the spokesperson is a male and F be the event that the spokesperson is over 35 years of age.    Since only one male is over 35 years of age,   We know, Therefore, the probability that the spokesperson will either be a male or over 35 years of age is .

7.(iv)  A and B are two events such that  \small P(A)=0.54, \small P(B)=0.69 and  \small P(A\cap B)=0.35 . Find  (iv) \small P(B\cap A{}')

Given, P(A) = 0.54, P(B) = 0.69, P(A  B) = 0.35 And, P(A  B) = 0.88 (iv) P(B  A') = P(B-A) = P(B) - P(A  B)  = 0.69 - 0.35 = 0.34  P(B  A') = 0.34

7.(iii)  A and B are two events such that  \small P(A)=0.54 ,  \small P(B)=0.69  and  \small P(A\cap B)=0.35. Find (iii) \small P(A\cap B{}')

Given, P(A) = 0.54, P(B) = 0.69, P(A  B) = 0.35 And, P(A  B) = 0.88 (iii) P(A  B') = P(A-B) = P(A) - P(A  B) = 0.54 - 0.35 = 0.19   P(A  B') = 0.19

7.(ii)  A and B are two events such that \small P(A)=0.54,  \small P(B)=0.69 and  \small P({A}\cap {B})=0.35. Find (ii)  \small P({A}'\cap {B}')   

Given,  And,  (ii)   [De Morgan’s law] So,   

7.(i)    A and B are two events such that  \small P(A)=0.54 , \small P(B)=0.69 and  \small P(A\cap B)=0.35 Find  (i)  \small P(A\cup B)

Given, P(A) = 0.54, P(B) = 0.69, P(A  B) = 0.35 (i) We know, P(A   B) = P(A) + P(B) - P(A  B)  P(A   B) = 0.54 + 0.69 - 0.35 = 0.88  P(A   B) = 0.88

6.  Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

Given, 3 letters are put in 3 envelopes. The number of ways of putting the 3 different letters randomly = 3! Number of ways that at least one of the 3 letters is in the correct envelope   = No. of ways that exactly 1 letter is in correct envelope +  No. of ways that 2 letters are in the correct envelope(The third is automatically placed correctly) = No. of ways that exactly 1 letter is in...

5.(b)  Out of  \small 100 students, two sections of \small 40 and  \small 60 are formed. If you and your friend are among the \small 100 students, what is the probability that

(b) you both enter the different sections?

Total number of students = 100 Let A and B be the two sections consisting of 40 and 60 students respectively. We found out in (a) that the probability that both students are in same section is  (b) Probability that both the students are in different section =   

5.(a)  Out of  \small 100  students, two sections of \small 40 and \small 60 are formed. If you and your friend are among the \small 100 students, what is the probability that

(a) you both enter the same section? 

Total number of students = 100 Let A and B be the two sections consisting of 40 and 60 students respectively. Number of ways of selecting 2 students out of 100 students.=  (a)  If both are in section A: Number of ways of selecting 40 students out of 100 =   (The remaining 60 will automatically be in section B!) Remaining 38 students are to be chosen out of (100-2 =) 98 students  Required...

4.(c)   In a certain lottery  \small 10,000  tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (c)  \small 10 tickets. 

Given, Total number of tickets sold = 10,000 Number of prizes awarded = 10 (c) If ten tickets are bought: Number of tickets not awarded = 10000 - 10 = 9990  P(not getting a prize) =   

4.(b)  In a certain lottery  \small 10,000  tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (b) two tickets 

Given, Total number of tickets sold = 10,000 Number of prizes awarded = 10 (b) If two tickets are bought: Number of tickets not awarded = 10000 - 10 = 9990  P(not getting a prize) = 

4.(a)  In a certain lottery  \small 10,000  tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket

Given, Total number of tickets sold = 10,000 Number of prizes awarded = 10 (a) If one ticket is bought, P(getting a prize) =     P(not getting a prize) = 1 - P(getting a prize)  

3.(iii) A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine

(iii) P(not 3) 

Total number of faces of a die = 6  (iii) Number of faces with number '3' = 1     P(not 3) = 1 - P(3) Therefore, required probability P(not 3) is   

3.(ii)  A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine

 (ii) P(1 or 3) 
 

Total number of faces of a die = 6  (ii) P (1 or 3) = P (not 2) = 1 − P (2)   Therefore, required probability P(1 or 3) is 0.5

3.(i)  A die has two faces each with number '1', three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine

(i) \small P(2)

Total number of faces of a die = 6  (i) Number of faces with number '2' = 3   Therefore, required probability P(2) is 0.5

2.   \small 4 cards are drawn from a well – shuffled deck of \small 52 cards. What is the probability of obtaining  \small 3 diamonds and one spade?

Total number of ways of drawing 4 cards from a deck of 52 cards =  We know that there are 13 diamonds and 13 spades in a deck. Now, Number of ways of drawing 3 diamonds and 1 spade =   Probability of obtaining 3 diamonds and 1 spade
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