Question

Form two digit numbers using the digit 0, 1, 2, 3, 4, 5 without repeating the digits. P is the event that the number so formed is even. Q is the event that the number so formed is divisible by 3. R is the event that the number so formed is greater than 50. S is the sample space.
Which of the options is correct?
(a) n(S)= 25, n(P)= 13, n(Q)= 9,n(R)= 5
(b) n(S)= 25, n (P)= 13, n(Q)= 9, n(R)=2
(c) n(S)= 25, n(P)= 13, n(Q)= 9,n(R)= 4
(d) n(S)= 25, n(P)= 13, n(Q)= 9,n(R)= 8

Answer 1

To solve the given question first we need to find the sample space, n(S) i.e. total possible outcomes. We can find it by adding all the possible outcomes of that event. For, example, to find n(P), since it is an event where numbers are even, we need to look at the numbers ending with 0, 2 and 4 only. Also, if we take n(Q), then we have to add all such two digit numbers that are divisible by 3, i.e. they should have their sum as a multiple of 3. We can do the same for events R and S.

Complete step-by-step answer:
We have been given the digits as 0, 1, 2, 3, 4, 5. It is also mentioned that we have to form two digit numbers using these digits, without repetition.
Now, to form two digit numbers, we know that 0 cannot come in the first place. The number of digits we have is 6, and excluding 0, we get it as 5. So, now let us check all the possible outcomes for this. We get that the first place can be filled with 5 digits and since there is no repetition allowed, the second place can be filled with another 5 digits. So, we can multiply them to get the sample space as shown below,
$n\left( S \right)=5\times 5=25$
Let us consider the event P now. It is given that it is the event that the number formed is an even number. We know that even numbers are divisible by 2. Also, we know that for a number to be divisible by 2, it must end with either 0, 2, 4, 6 or 8. Since the given digits are 0, 1, 2, 3, 4 and 5, the possible cases would be 0, 2 and 4.
Considering the possible cases, we will get that there can be five numbers with 0 at the end; there can be 4 numbers with 2 at the end (excluding the digit zero); there can be 4 numbers with 4 at the end (excluding digit zero). So, now adding all the possibilities, we get,
$n\left( P \right)=5+4+4=13$
Now for the event Q, we must look at all possible two digit numbers divisible by 3. Consider 1 in the first place of the two digit number, then 12 and 15 are the possible combinations (since 1+2 and 1+5 are multiples of 3). Similarly for 2 in first place 21, 24 are possible two digit numbers; 30 for 3; 41, 45 for 4; 51,54 for 5.Hence 12, 15, 21, 24, 30, 42, 45, 51, 54 are the possible two digit numbers which are divisible by 3.
n(Q)= 9
Now the two digit numbers greater than 50 will have 5 in their first place and numbers 1, 2, 3, 4 in second place (excluding 5 since repetition is not allowed). Thus 51, 52, 53, 54 are the possible two-digit numbers greater than 50.
n(R)= 4
So, The correct option is C

Note: Never consider 0 in the first place of the two-digit number. Also since the digits are not repeating the first and second place should be filled with two different numbers. Suppose repetition was allowed, then we would have had the sample space as $n\left( S \right)=5\times 6=30$. While finding sample space for event Q, students must write down each multiple of 3 in order, so that nothing is missed. Sometimes students miss to read the first words of the question, i.e. “two digit number” and this will further lead to wrong computations. So, it is advised to read the question carefully and then proceed.