Question

Q10.
A three digit number is to be formed using the digits 3, 4, 7, 8 and 2 without repetition. The probability that it is an odd
number is
(A ) 2/5
(B) 1/5
(C) 4/5
(D) 3/5

Answer 1

Answer:

(A)2/5 is the correct answer...

hope it helps,,,,

Answer 2

The probability that it is an odd  number is $\dfrac{2}{5}$ .

Explanation:

Total digits ( 3, 4, 7, 8 and 2) :5

If repetition is not allowed , then we use permutations.

Permutation of r things selecting from n things = $^nP_r=\dfrac{n!}{(n-r)!}$

If we make a three digit number is to be formed using the digits 3, 4, 7, 8 and 2 without repetition , then the total different numbers formed = $^5P_3=\dfrac{5!}{(5-3)!}=\dfrac{5\times4\times3\times2!}{2!} =60$

To make odd number , the last digit has 2 choice (3,7).

Fix 3 on last digit , the number of ways to make odd numbers = 3 x4x1 =12

Fix 7 on last digit , the number of ways to make odd numbers = 3 x4x1 =12

Total odd numbers can be formed = 12+12=24

Now , the probability that it is an odd  number would be $\dfrac{\text{No. of odd numbers}}{\text{Total numbers formed}}$

$=\dfrac{24}{60}=\dfrac{2}{5}$

Hence, the probability that it is an odd  number is $\dfrac{2}{5}$ .

# Learn more :

Two digit number are to made using the digits 3,5,6 and 8, without repetition of digits . Find the probability of the following events( 1) the number is odd. (2) the number is divisible by 9.

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