Chapter 33 Binomial Distribution Exercise Ex. 33.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Required Probability =4547 over 8192Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Also, find the mean and variance of this distribution.Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

                                = 0.0256Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solution 47

Question 48

Solution 48

Question 49

Solution 49

Question 50

Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.Solution 50

Question 51

A die is thrown 5 times. Find the probability that an odd number will come up exactly three times.Solution 51

Question 52

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?Solution 52

Question 53

A factory produces bulbs. The probability that one bulb is defective is   and they are packed in boxes of 10. From a single box, find the probability that

i. none of the bulbs is defective.

ii. exactly two bulls are defective.

iii. more than 8 bulbs work properly.Solution 53

Note: Answer given in the book is incorrect.

Chapter 33 Binomial Distribution Exercise Ex. 33.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

From a lot of 15 bulbs which include 5 defective, sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence, find the mean of the distribution.Solution 22

Out of 15 bulbs 5 are defective.

begin mathsize 12px style table attributes columnalign left end attributes row cell text Hence ,  the   probability   that   the   drawn   bulb   is   defective   is end text end cell row cell text P end text left parenthesis text Defective end text right parenthesis equals 5 over 15 equals 1 third end cell row cell text P end text left parenthesis text Not   defective end text right parenthesis equals 10 over 15 equals 2 over 3 end cell row cell text Let   X   denote   the   number   of   defective   bulbs   out   of   4. end text end cell row cell text Then ,  X   follows   binomial   distribution   with   end text end cell row cell straight n equals 4 comma text   end text straight p equals 1 third text   and   end text straight q equals 2 over 3 text   such   that end text end cell row cell straight P left parenthesis straight X equals straight r right parenthesis equals straight C presuperscript 4 subscript straight r open parentheses 1 third close parentheses to the power of straight r open parentheses 2 over 3 close parentheses to the power of 4 minus straight r end exponent semicolon straight r equals 0 comma 1 comma 2 comma 3 comma 4 end cell row cell text Mean end text equals sum from straight r equals 0 to 4 of rP left parenthesis straight r right parenthesis equals 1 cross times straight C presuperscript 4 subscript 1 open parentheses 1 third close parentheses open parentheses 2 over 3 close parentheses cubed plus 2 cross times straight C presuperscript 4 subscript 2 open parentheses 1 third close parentheses squared open parentheses 2 over 3 close parentheses squared end cell row cell plus 3 cross times straight C presuperscript 4 subscript 3 open parentheses 1 third close parentheses cubed open parentheses 2 over 3 close parentheses plus 4 cross times straight C presuperscript 4 subscript 4 open parentheses 1 third close parentheses to the power of 4 open parentheses 2 over 3 close parentheses to the power of 0 end cell row cell equals 32 over 81 plus 48 over 81 plus 24 over 81 plus 4 over 81 equals 108 over 81 equals 4 over 3 end cell end table end style

Question 23

A die is thrown three times. Let X be’ the number of twos seen’. Find the expectation of X.Solution 23

Question 24

A die is thrown twice. A ‘success’ is getting an even number on a toss. Find the variance of number of successes.Solution 24

Question 25

Three cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability of the number spades. Hence, find the mean of the distribution.Solution 25


Discover more from EduGrown School

Subscribe to get the latest posts sent to your email.