Question

3 digit numbers are made using 4,5,9 without repetition. If a number among is selected at random. What will be the probability of getting i) Multiple of 5 ii) Multiple of 9 iii) will end with 9

Answer 1

Answer: There can be three combination of the numbers 4,5,9

459,495,594,549,945,954

Therefore total number of possible outcomes=6

I)multiple of 5-495,945

No. Of favourable outcomes=2

P(getting multiple of 5)=2/6=1/3

II)multiple of 9

All the numbers are multiple of 9

No. Of favourable outcomes=6

P(getting a multiple of 9)=6/6=1

III)will end with zero

Nos. Ending with 9=549,459

No. Of favourable outcomes=2

P(getting a no. Which ends with zero)=2/6=1/3

I hope it helps

I think the other answer given by someone else given is wrong because all the nos. Are divisible by 9

Step-by-step explanation:

Answer 2

Total number of three digit numbers are :

459,  495, 549, 594, 945, 954 = 6

(i) P(multiple of 5) = $\bf\huge\frac{2}{6}$ = $\bf\huge\frac{1}{3}$

(ii) P(multiple of 9) = $\bf\huge\frac{6}{6}$ = 1

(iii) P(ending with 9) =  $\bf\huge\frac{2}{6}$ = $\bf\huge\frac{1}{3}$