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Q1. The digits 1,2,3,....,9 are written in random order to form a 9 digit number.find the probability that this number will be divisible by 2?

Solution

These 9 digits can be arranged in 9! ways. for the number to be divisible by 2 it should be even i.e. the last digit should be either 2,4,6,8. which can be chosen in 4 ways then the remaining 8 digits can be arranged in 8! ways. So the required probability is=fraction numerator 4.8 factorial over denominator 9 factorial end fraction equals 4 over 9
Q2. Out of 13 applicants for a job, there are 5 women and 8 men.It is desired to select 2 persons for the job. The probability that atleast one of the selected person will be woman?

Solution

The probability that none of the women will be selected=The probability of both men selected The probability of both men selected=fraction numerator C presuperscript 8 subscript 2 over denominator C presuperscript 13 subscript 2 end fraction equals fraction numerator 8.7 over denominator 13.12 end fraction equals 14 over 39 The probability of atleast one woman will be selected=1-14 over 39 equals 25 over 39
Q3. In a class 40% of the students offered Physics 20% offered Chemistry and 5%offered both.If a student is selected at random then find the probability that he has offered Physics or Chemistry only?

Solution

text Let A=student offered Physics⇒P(A)= end text 40 over 100 text B=student offered Chemistry⇒P(B)= end text 20 over 100 text probability that student offered both the subjects is P(A∩B)= end text 5 over 100 text Then the required probability is P(A∪B)=P(A)+P(B)-P(A∩B) end text space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 40 over 100 plus 20 over 100 minus 5 over 100 equals 55 over 100
Q4. What do you mean by mutually exclusive and exhaustive events?

Solution

Let S be a sample space associated with a random experiment. Let E subscript 1 comma E subscript 2 comma........ comma E subscript n space b e space t h e space s u b s e t s space o f space S space s u c h space t h a t 1 right parenthesis E subscript i intersection E subscript j equals empty set space f o r space i not equal to j 2 right parenthesis E subscript 1 union E subscript 2....... union E subscript n equals S comma space t h e n space t h e space s e t space o f space e v e n t s space E subscript 1 comma space E subscript 2 comma....... E subscript n space a r e space s a i d space t o space b e space m u t u a l l y space e x c l u s i v e space a n d space e x h a u s t i v e space s y s t e m space o f space e v e n t s.
Q5. A natural number is chosen at random from first 100 natural numbers.What is the probability that the number chosen is divisible by  3 or 5?

Solution

text Let A={3,6,9,....,99} and B={5,10,15,,,,,,100} and A∩B={15,30,.....,90} end text n left parenthesis A right parenthesis equals 33 space a n d space n left parenthesis B right parenthesis equals 20 space a n d space n left parenthesis A intersection B right parenthesis equals 6 t h e space r e q u i r e d space p r o b a b i l i t y space i s space P left parenthesis A union B right parenthesis equals P left parenthesis A right parenthesis plus P left parenthesis B right parenthesis minus P left parenthesis A intersection B right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 33 over 100 plus 20 over 100 minus 6 over 100 equals 47 over 100


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