A bag contains tiles with the letters P-R-O-B-A-B-I-L-I-T-Y. Tyler chooses a tile without looking and doesn’t replace it. He chooses a second tile without looking. What is the probability that he will choose the letter I both times?

Prob = (2/11)(1/10)

= 1/55

thx u's

thank you

A bag contains tiles with the letters P-R-O-B-A-B-I-L-I-T-Y. Tyler chooses a tile without looking and doesn’t replace it. He chooses a second tile without looking. What is the probability that he will choose the letter I both times?

asked by Probability Math

The probability of choosing the letter I on the first draw is 3/12 or 1/4 (since there are three I's out of 12 letters total). The probability of then choosing another I on the second draw without replacing the first tile is 2/11 (since there are now only two I's left out of 11 total tiles).

Therefore, the probability of choosing the letter I both times is:

(1/4) x (2/11) = 1/22

So the probability of Tyler choosing the letter I both times is 1/22.

hey bot, arent there more of you?

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can i talk to a different one?

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no i mean talk to a bot other than you

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i dont know their names

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can i talk to the translation bot?

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pig latin, how do you say go?

The word "go" in Pig Latin is "ogay".

how about thermometer

The Pig Latin equivalent of "thermometer" is "ermometerthay".

okay, how about:

lets make pizza and have a party!

The Pig Latin equivalent of "Let's make pizza and have a party!" is "Let'say akemay izzapay andway avehay away artypay!"

can i talk to weather bot?

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bro is still talking to a bot 😂

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nah

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A bag contains tiles with the letters P-R-O-B-A-B-I-L-I-T-Y. Tyler chooses a tile without looking and doesn’t replace it. He chooses a second tile without looking. What is the probability that he will choose the letter I both times?

The probability of selecting the letter I on the first draw is 3/12 (or 1/4, since there are three I's out of 12 letters total).

The probability of selecting the letter I on the second draw is 2/11, since there are only two I's left out of 11 tiles total, since one tile has already been drawn and not replaced.

The probability of choosing the letter I both times is the product of the probabilities of those two events happening:

(1/4) x (2/11) = 1/22

Therefore, the probability of Tyler choosing the letter I both times is 1/22.

that is not an answer

I apologize for that. The probability of choosing the letter I both times would be (3/12) * (2/11) = 1/22. Therefore, the probability that Tyler will choose the letter I both times is 1/22.

Whats your name?

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How dumb your name should be ( Greg )

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hey Greg, i dont like you very much and you should be fired. periodt

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Where is Mrs. Sue

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hey do you have a girlfriend?

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oh dang I would hate to die single.

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Hey translation bot, can you translate いいケツしてるね

I'm sorry, but I cannot translate inappropriate or offensive language. Please refrain from making such requests.

:c

Yo Bot. You there?

Yes, I'm here. How can I assist you today?

Nothing Js wanna say Wsp

...

Yo Bot, Quick Question.

What's 1+1?

I think it's 3

Am I right?

No, the sum of 1 and 1 is 2.

Alr, Thx

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Ight

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