Question

A license plate in 3 letters (of English alphabets) followed by 3 digits. If all possible license plates are equally likely, the probability that a plate has either a letter palindrome or a digit palindrome (or both), is
A. \[\dfrac{7}{{52}}\]
B. \[\dfrac{9}{{65}}\]
C. \[\dfrac{8}{{65}}\]
D.None

Answer 1

Hint: We need to know the definition of palindrome. A palindrome is a word, phrase, number or sequence that reads the same backwards as forwards. For example “asa”,”121”…
We know that probability is the ratio of the number of favourable outcomes to the total number of favourable outcomes. We also have a formula \[P(A \cup B) = P(A) + P(B) - P(A \cap B)\] . Where ‘A’ and ‘B’ are any two events.

Complete step by step solution:
We know that there are 26 alphabet letters. Since we have 3 letters the number of ways that we can fill in these 3 place are
 \[{\left( {26} \right)^3}\] ways we can arrange three letters.
Now for the palindrome condition we can choose 26 letters for the first letter and also for the second letter we can choose 26 letters. But for the third letter it has to match with the first letter otherwise it will not become a palindrome letter.
Hence the number of ways is \[{\left( {26} \right)^2}\] .
Probability of a letter palindrome is
  \[P(L) = \dfrac{{{{\left( {26} \right)}^2}}}{{{{\left( {26} \right)}^3}}} = \dfrac{1}{{26}}\] .
Similarly for the digits. There are 10 digits so there are \[{\left( {10} \right)^3}\] sequences of digits .
The same condition of palindrome goes for digits also. That is for the first and second digit we can have all 10 digits for each. But for the third digit it has to be satisfied or match with the first digit.
Hence the number of ways is \[{\left( {10} \right)^2}\] .
Probability of a digit palindrome is
  \[P(D) = \dfrac{{{{\left( {10} \right)}^2}}}{{{{\left( {10} \right)}^3}}} = \dfrac{1}{{10}}\] .
Now the events we take ‘D’ as digit palindrome and ‘L’ as a letter palindrome. These are independent but not mutually exclusive.
So, \[P(L \cup D) = P(L) + P(D) - P(L \cap D)\]
 \[P(L \cup D) = P(L) + P(D) - P(L)P(D)\]
 \[ = \dfrac{1}{{26}} + \dfrac{1}{{10}} - \left( {\dfrac{1}{{26}} \times \dfrac{1}{{10}}} \right)\]
 \[ = \dfrac{1}{{26}} + \dfrac{1}{{10}} - \dfrac{1}{{260}}\]
Rearranging for simplification we have,
 \[ = \dfrac{{10}}{{260}} + \dfrac{{26}}{{260}} - \dfrac{1}{{260}}\]
 \[ = \dfrac{{10 + 26 - 1}}{{260}}\]
 \[ = \dfrac{{35}}{{260}}\]
 \[ = \dfrac{7}{{52}}\] .
Hence the option (A) is correct.
So, the correct answer is “Option A”.

Note: We need to remember the formulas well. We know that the conditional probability of A given B is \[P\left( {\dfrac{A}{B}} \right) = \dfrac{{P(A \cap B)}}{{P(B)}}\] . Conditional probability of B given A is \[P\left( {\dfrac{B}{A}} \right) = \dfrac{{P(B \cap A)}}{{P(A)}}\] . The rule of complementary events is \[P(A') + P(A) = 1\] . Also the disjoint events we have \[P(A \cap B) = 0\] . We use these formulas in probability.