Independent and dependent events: tossing coin, rolling die

peggyskold

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Jan 30, 2009
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Hello--back again and lost again as well. I cannot seem to get my head around this.

Tossing a coin and getting a head (A), and then rolling a six-sided die and obtaining a 6 (B).

The solution is P(B\A) = 1/6 and P(B) = 1/6

I get that the event of rolling the die and getting a six does not affect the probability of flipping a coin and getting a head, so the events are independent. However; I don't get the numbers? Help anyone? Thanks
 
Re: Independent and dependent events

peggyskold said:
Hello--back again and lost again as well. I cannot seem to get my head around this.
Tossing a coin and getting a head (A), and then rolling a six-sided die and obtaining a 6 (B).
The solution is P(B\A) = 1/6 and P(B) = 1/6 I get that the event of rolling the die and getting a six does not affect the probability of flipping a coin and getting a head, so the events are independent. However; I don't get the numbers? Help anyone? Thanks

Do you understand - why P(B) = 1/6?
 
Re: Independent and dependent events

Yes I understand that you have six possible outcomes to roll a 6 on the die and 1roll ,so 1/6. Correct?
 
Re: Independent and dependent events

peggyskold said:
Yes I understand that you have six possible outcomes to roll a 6 on the die and I roll so 1/6. Correct?

Correct ....

Then

P(B/A) means probability of "B happening" given "A happend" - you know that.

Since A and B are independent

P(B/A) = P(B) = 1/6
 
Re: Independent and dependent events

I'm sorry--still cant quite get it. So, what did we do with the probability of the coin which would be 1/2 correct?
 
Re: Independent and dependent events

peggyskold said:
I'm sorry--still cant quite get it. So, what did we do with the probability of the coin which would be 1/2 correct?
We throw the coin away. It no longer has any effect of the question
 
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