Question

33. A number is chosen at random from 1 to 100. The probability of getting a number which is either multiple of 2 or multiple of 5 but not both is (a) 3/5 (b) 7/10 (c) 4/5 (d) 3/10​

Answer 1

Answer:

A)3/5

Step-by-step explanation:

Total sample space=100

no of 2 multiples=100/2=50

no of 5 multiples=100/5=20

LCM of (2,5)=10

multiples of 10 => 10,20,30,40,50,60,70,80,90,100 (all have 2 and 5 as factors)

∴ 10 multiples

∴Probability= $\frac{outcomes favorable to the event }{total number of outcomes}$

=$\frac{(No of 2 multiples + No of 5 multiples)-(both 2 and 5 factor no)}{100}$

=$\frac{(50+20)-10}{100}$

=$\frac{60}{100}$

after simplification we get,

=$\frac{3}{5}$

Answer 2

Given :  A number is chosen at random from 1 to 100.

To Find :  The probability of getting a number which is either multiple of 2 or multiple of 5 but not both

(a) 3/5 (b) 7/10 (c) 4/5 (d) 3/10​

Solution:

1 to 100.

n(S) = 100

Multiples of 2

2  ,4 , 6 , ... , 98 , 100

50 numbers

Multiples of 5

5 , 10 , 15 ,... , 95 , 100

20  numbers

Multiples of both  means multiples of LCM (2 , 5) = 10

Multiples of 10

10 , 20 , ... , 90 , 100

10 Numbers

number which is either multiple of 2 or multiple of 5 but not both  

= Multiple of 2 only + Multiple of 5 only

= ( multiple of 2  - Multiple of 10)  + (multiple of 5 - Multiple of 10)

= (50 - 10) + (20 - 10)

= 40 + 10

= 50

E = Event of  getting a number which is either multiple of 2 or multiple of 5 but not both

n(E) = 50

P(E) = n(E)/n(S)

Probability = 50/100 = 1/2

None of the given option is correct

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