Question

A locker requires a three-digit code to open the lock. The code must contain one letter and two numbers, and no letter or number can be repeated. You can choose from among four letters, A, B, C, and D, and two numbers, 5 and 6.
The size of the sample space is (8, 16, 20, 24)

If a code is chosen at random, the probability that it has a letter that immediately follows an odd number is (1/8, 1/6, 1/3, 2/5, 2/3)

If a code is chosen at random, the probability that D is in the code but is not in the first position is (1/8, 1/6, 1/3, 2/5, 2/3)

Answer 1

So it can be:

Letter Number Number (and any permutation: Number Letter Number, or Letter Number Number). Three times one case.

For Letter Number Number, there are 4 letters and two numbers. Since numbers cannot be repeated, both of them must be used, because there must be two numbers. So, indeed:

4 * 2 * 1 = 8, but any permutation, so 8 * 3 = 24 is the sample size.

Second question: Must be,

5 Letter 6, so 4 cases: P= 4/24 = 1/6

Third question. May be:

5 D 6, 6 D 5, 5 6 D, or 6 5 D

P = 4/24 =1/6, again!

Answer 2

Answer with explanation:

→→(A):

Number of digits required to make or open a lock = 3→ [1 letter +2 number]

Number of letters from which we have to choose , 1 Letter = 4={A,B,C,D}

Number of digits from which we have to choose a digit =2={5,6}

Size of the Sample Space

                          =Choosing a letter from 4 letters × Placing 2 numbers that is 5 and 6.

={(A,5,6),(B,5,6),(C,5,6),(D,5,6)}

→Each of the 3 digit lock has ,3! or

                                 $_{3}^{3}\textrm{P}$

=6, distinct arrangement.

So, total Size of sample Space=6 × 4=24 ⇒→Option D

→→(B):

There is one odd number, that is, 5,so the odd number must come in between the three digit code, Out of 24, sample space .

So, Favorable Outcome={(6 5 A),(6 5 B),(6 5 C),(6 5 D)}=4  

Required Probability

         $=\frac{\text{Total Favorable Outcome}}{\text{Total Possible Outcome}}\\\\=\frac{4}{24}\\\\=\frac{1}{6}$

Option B

$=\frac{1}{6}$

→→(C):

Out of 24 , three digit codes, number of codes in which D is in the code but is not in the first position ={(5,D,6),(6,D,5),(5,6,D),(6,5,D)}=4  

Required Probability

         $=\frac{\text{Total Favorable Outcome}}{\text{Total Possible Outcome}}\\\\=\frac{4}{24}\\\\=\frac{1}{6}$

Option B

$=\frac{1}{6}$