Question

13. A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the
digits is allowed. Find the probability that the number so formed is a -
(1) prime number (2) multiple of 4
(3) multiple of 11.
​

Answer 1

Given

.A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the

digits is allowed. Find the probability that the number so formed is a -

(1) prime number (2) multiple of 4

(3) multiple of 11.

​Answer

Numbers are:

10,11,12,13,14,20,21,22,23,24,30,31,32,33,34,40,41,42,43,44

1) Prime number

11,13,23,29,31,37,41,43

P=  

8/20 fraction

2)Multiple of 4

12,16,20,24,28,32,36,40,44

P= 20/9(fraction)

3)Multiple of 11

11,22,33,44

P= 4/20 fraction

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#ANGRY ARMY

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Answer 2

Answer:

Given : A two digit number is to be formed from the digits 0, 1, 2, 3, 4. Repetition of the digits is allowed.

To Find: Find the probability that the number so formed is a -

(1) prime number

(2) multiple of 4

(3) multiple of 11.

Solution:

To find total number of possible outcomes we will use combination

^nC_r =\frac{n!}{r!(n-r)!}

n

C

r

=

r!(n−r)!

n!

We are given that repetition is allowed.

Now total no. of digite are {0,1,2,3,4} = 5

Out which we are supposed to make 2 digit number

So, n = 5

r = 1

^6C_1 =\frac{6!}{1!(6-1)!}

6

C

1

=

1!(6−1)!

6!

^6C_1 =\frac{6!}{1!(5)!}

6

C

1

=

1!(5)!

6!

^6C_1 =6

6

C

1

=6

Now for tens place we have selected the digit

Now for ones Place since the repetition is allowed .So, n will remain 6

So, n = 5

r= 1

^6C_1 =\frac{6!}{1!(6-1)!}

6

C

1

=

1!(6−1)!

6!

^6C_1 =\frac{6!}{1!(5)!}

6

C

1

=

1!(5)!

6!

^6C_1 =6

6

C

1

=6

So, total possible outcomes = 6*6 = 36

1)prime number

Total outcomes = 36

Prime numbers can be made from given digits = {11,13,23,31,41,43} = 6

So, probability that the number so formed is a prime = \frac{6}{36} =\frac{1}{6}

36

6

=

6

1

(2) multiple of 4

Multiple of 4 can be made from given digits = {12,20,24,32,40,44} = 6

So, probability that the number so formed is a Multiple of 4 = \frac{6}{36} =\frac{1}{6}

36

6

=

6

1

3)multiple of 11.

Multiple of 11 can be made from given digits = {11,22,33,44} = 4

So, probability that the number so formed is a Multiple of 11 = \frac{4}{36} =\frac{1}{9}

36

4

=

9

1

Brainliest answer