Help: Conditional Probability

[caffinatedloz]

Hello! Our teacher set us a bunch of conditional probability for homework over the holidays and hasn't given us the solutions. I was just wondering if someone could double check I did this question right.

Thanks!!

Two cards are drawn in succession from a standard 52-card deck. What is the probability that both cards are aces if the cards are drawn without replacement?

I did (4/52) * (3/51)

= 0.45%


Also, this question!

You roll two dice. The first die shows a ONE and the other die rolls under the table and you cannot see it. Now, what is the probability that both die show ONE?

Would the probability be 1/6? It seems too simple so I am probably missing something. Ahahaha.

Mod edit: Merged double post

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Also, this question!

You roll two dice. The first die shows a ONE and the other die rolls under the table and you cannot see it. Now, what is the probability that both die show ONE?

Would the probability be 1/6? It seems too simple so I am probably missing something. Ahahaha.

Hey there, I am yr 9 and I do a accelerated program in my school. We just finished probability and logarithms/surds this semester so I might be able to help u.

For ur first one :

Hello! Our teacher set us a bunch of conditional probability for homework over the holidays and hasn't given us the solutions. I was just wondering if someone could double check I did this question right.

Thanks!!

Two cards are drawn in succession from a standard 52-card deck. What is the probability that both cards are aces if the cards are drawn without replacement?

I did (4/52) * (3/51)

= 0.45%

ur correct . There are 4 aces in a die and the probability of picking one is at first, 4 out of 52 (52 cards in a deck of cards). As the next does not require a replacement there would then be 51 cards (assuming that the first card picked is an ace). so the chance will be  3 out of 51.
Now we multiply these to get the final result:

4/52 * 3/51 = 0.0045.. or 0.45%


U r also right for the second one:

there are 6 faces to a die. The probability of a die showing ONE is 1/6. (We will not need to find the probability of both die as the first die is already given). Therefore the probability of both die showing ONE is  1/6.

Hope that helped