OR probabilities

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  • Thread starter little neutrino
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I am still sure that I did the calculations right, but let's go with your approach. P(A) = 75/676 because in the first two cases we have 2 letters the same, and in the third case we have 3 letters the same.P(B) = 1/9 because we have exactly one digit the same in the first two cases, and no digits the same in the third case.P(A \wedge B) = 25/2028 because we have 2 letters the same AND 2 digits the same in the third case.So using the formula, we get P(A\ xor\ B) = (75/676) + (1/9) - 2(25/
  • #1
little neutrino
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A code is generated consisting of 3 letters followed by 2 digits. Each of the 3 letters generated is equally likely to be any of the 26 letters of the alphabet. Each of the 2 digits generated is equally likely to be any of the nine digits 1-9. Find the probability that a randomly chosen code has exactly 2 letters the same or 2 digits the same, but not both.

Attempted solution:

P(exactly 2 digits same) = (263*9)/(263*92) = 1/9
P(exactly 2 letters same) = (26*25*3*92)/(263*92) = 75/676
P(2 digits same AND 2 letters same) = (26*25*3*9)/(263*92) = 25/2028
Required probability = 1/9 + 75/676 - 25/2028 = 0.210

But the correct answer is 0.197. Can I ask where did I go wrong? I can't figure it out...
Thanks!
 
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  • #2
little neutrino said:
A code is generated consisting of 3 letters followed by 2 digits. Each of the 3 letters generated is equally likely to be any of the 26 letters of the alphabet. Each of the 2 digits generated is equally likely to be any of the nine digits 1-9. Find the probability that a randomly chosen code has exactly 2 letters the same or 2 digits the same, but not both.

Attempted solution:

P(exactly 2 digits same) = (263*9)/(263*92) = 1/9
P(exactly 2 letters same) = (26*25*3*92)/(263*92) = 75/676
P(2 digits same AND 2 letters same) = (26*25*3*9)/(263*92) = 25/2028
Required probability = 1/9 + 75/676 - 25/2028 = 0.210

But the correct answer is 0.197. Can I ask where did I go wrong? I can't figure it out...
Thanks!

Your calculation for 2 letters is the probability that at LEAST 2 letters is the same. You have to subtract off the probability that all 3 letters are the same to get the probability that exactly 2 letters are the same.
 
  • #3
One other possible mistake. Exactly how many ways are there to have 2 items the same in a set of 3?
 
  • #4
stevendaryl said:
Your calculation for 2 letters is the probability that at LEAST 2 letters is the same. You have to subtract off the probability that all 3 letters are the same to get the probability that exactly 2 letters are the same.

But why is that so?
__ __ __ __ __
For two letters to be the same == 26 choices
For the last letter, since it can't be the same as the two letters that are the same == 26 - 1 = 25 choices
No. of ways to arrange 3 letters, when 2 are the same == 3!/2! = 3
Thus no. of ways == 26 * 25 * 3 * 92

Why did I calculate the probability that at least 2 letters are the same? Could you elaborate? Sorry kind of confused.
 
  • #5
little neutrino said:
No. of ways to arrange 3 letters, when 2 are the same == 3!/2! = 3
Yup I'm dumb sorry :D
 
  • #6
Hmm. My criticism was mistaken, but here's a suggestion: Instead of computing probabilities for the various possibilities, try just computing

[itex]N_L =[/itex] the number of ways to have exactly 2 letters the same, and both digits different.
[itex]N_D =[/itex] the number of ways to have 3 or 0 letters the same, and both digits the same.

Then the probability is [itex]\frac{N_L + N_D}{26^3 \cdot 9^3}[/itex]

When I did it that way, I got the expected answer.
 
  • #7
Okay, I figured out what you did wrong:

You used [itex]P(A \vee B) = P(A) + P(B) - P(A \wedge B)[/itex] where [itex]A[/itex] = "exactly 2 letters the same" and [itex]B[/itex] = "both digits the same". But they aren't asking for [itex]A \vee B[/itex], they're asking for [itex]A\ xor\ B[/itex] ([itex]xor[/itex] = exclusive or). The formula for that is:

[itex]P(A\ xor\ B) = P(A) + P(B) - 2P(A \wedge B)[/itex]
 

What is the "OR" probability rule?

The "OR" probability rule states that the probability of either event A or event B occurring is equal to the sum of their individual probabilities, minus the probability of both events occurring together.

How is the "OR" probability calculated?

The "OR" probability can be calculated by adding the individual probabilities of event A and event B, and then subtracting the probability of both events occurring together.

What is the difference between "OR" and "AND" probabilities?

The "OR" probability represents the likelihood of either event A or event B occurring, while the "AND" probability represents the likelihood of both events occurring together.

Can the "OR" probability be greater than 1?

Yes, the "OR" probability can be greater than 1 if the individual probabilities of event A and event B are both greater than 0.5 and the probability of both events occurring together is also greater than 0.5.

How is the "OR" probability used in real life?

The "OR" probability is commonly used in risk assessment and decision making, as it allows for the consideration of multiple possible outcomes when making a decision or evaluating a situation.

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