Class 12-commerce RD SHARMA Solutions Maths Chapter 33: Binomial Distribution

Ex. 33.1

Ex. 33.2

MCQ

Ex. 33VSAQ

Binomial Distribution Exercise Ex. 33.1

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Required Probability = $4547 over 8192$

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

= 0.0256

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Note: Answer given in the book is incorrect.

Binomial Distribution Exercise Ex. 33.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

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$