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The number plates consists of 3 letters of the English alphabet follow 04 Mar 2021, 10:00
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The number plates consists of 3 letters of the English alphabet followed by 3 digits. What is the probability that a randomly picked number plate contain 3 same letters followed by 3 same digits ?

A. 1/260
B. 1/33,800
C. 1/43,200
D. 1/67,600
E. 1/135.200
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D
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The number plates consists of 3 letters of the English alphabet follow 04 Mar 2021, 12:47
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Given: The number plates consists of 3 letters of the English alphabet followed by 3 digits.
Asked: What is the probability that a randomly picked number plate contain 3 same letters followed by 3 same digits ?

Total number of ways for choosing the number plate = 26^3*10^3 = 17576000
Favorable number of ways for choosing the same number plate = 1
Probability = 1/17576000

A. 1/260
B. 1/33,800
C. 1/43,200
D. 1/67,600
E. 1/135.200

Bunuel
Please check what is the mistake in the above solution.
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The number plates consists of 3 letters of the English alphabet follow 04 Mar 2021, 13:26
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Kinshook
Given: The number plates consists of 3 letters of the English alphabet followed by 3 digits.
Asked: What is the probability that a randomly picked number plate contain 3 same letters followed by 3 same digits ?

Total number of ways for choosing the number plate = 26^3*10^3 = 17576000
Favorable number of ways for choosing the same number plate = 1
Probability = 1/17576000

A. 1/260
B. 1/33,800
C. 1/43,200
D. 1/67,600
E. 1/135.200

Bunuel
Please check what is the mistake in the above solution.
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Number of ways of making a number plate with 3 same letters followed by 3 same digits=26*1*1*10*1*1

So probability of picking a number plate containing 3 same letters followed by 3 same digits = (26*1*1*10*1*1)/(26*26*26*10*10*10)=1/67600

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The number plates consists of 3 letters of the English alphabet follow 08 Aug 2024, 04:37
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The probability of all alphabets same:

\(\frac{26}{26}*\frac{1}{26}*\frac{1}{26}\)

The first alphabet can be any letter so the probability is \(\frac{26}{26} = 1\). For the second letter to be the same as first, the probability is \(\frac{1}{26}\), similarly for third letter to be same too, we multiply again by \(\frac{1}{26}\).

For digits:

\(\frac{10}{10}*\frac{1}{10}*\frac{1}{10}\)

Same logic as above. We have 10 total choices for 0-9 digits.

Now multiply both together to find probability of all of these events occurring together:­

\((\frac{26}{26}*\frac{1}{26}*\frac{1}{26})(\frac{10}{10}*\frac{1}{10}*\frac{1}{10}) = \frac{1}{67600}\)­
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The number plates consists of 3 letters of the English alphabet follow 15 Jan 2026, 15:07
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Deconstructing the Question
Structure of the plate: [3 Letters] followed by [3 Digits].
Total Letters available = 26.
Total Digits available = 10.

Condition: "3 same letters" AND "3 same digits".

Step 1: Total Possible Outcomes
Since repetition is allowed (implied for license plates):
- Ways to choose 3 letters = \(26 \times 26 \times 26 = 26^3\)
- Ways to choose 3 digits = \(10 \times 10 \times 10 = 10^3\)
Total Outcomes = \(26^3 \times 10^3\).

Step 2: Favorable Outcomes
- Ways to have 3 same letters (AAA, BBB... ZZZ) = \(26\).
- Ways to have 3 same digits (000, 111... 999) = \(10\).
Favorable Outcomes = \(26 \times 10\).

Step 3: Calculate Probability
\(P = \frac{\text{Favorable}}{\text{Total}}\)
\(P = \frac{26 \times 10}{26^3 \times 10^3}\)

Simplify by canceling one 26 and one 10:
\(P = \frac{1}{26^2 \times 10^2}\)
\(P = \frac{1}{676 \times 100}\)
\(P = \frac{1}{67600}\)

Answer: D
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The number plates consists of 3 letters of the English alphabet follow 15 Jan 2026, 21:24
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The number plates consists of 3 letters of the English alphabet followed by 3 digits. What is the probability that a randomly picked number plate contain 3 same letters followed by 3 same digits ?

The correct option should be Option D. 1/67,600, and here's why:

Let any such number plate be visually represented as A1 A2 A3 D1 D2 D3 (where A1, A2, A3 represent 3 letters of the English aphabet and D1, D2, D3 represent the 3 digits present in the number plate).

We are required to find out the probability that a number plate contains A A A D D D (A represents the same letter and D represents the same digit)

Now, the total ways of forming such number plates = 26C1*1*1*10C1*1*1
Reasoning: The common letter can be selected in a total of 26C1 or 26 ways, this selected letter would be present at the 2nd and 3rd positions as well, hence there's only 1 way to select the 2nd 3rd alphabet. Now, for selecting the digits, the common digit can be selected in a total of 10C1 or 10 ways. This selected digit would present at the 5th and the 6th positions as well, hence we're left with only 1 way to select the 5th and 6th digits.

The overall total number of ways of forming the number plates without any constraint = 26C1*26C1*26C1*10C1*10C1*10C1 = 26*26*26*10*10*10

Therefore, required probability = (26*10)/(26*26*26*10*10*10) = 1/67,600

Hence, option (D) is the correct answer.
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