SOLUTION: What is the probability of drawing one 5 from a pack of cards if you are dealt 4 cards? And
If 3 dice are rolled what is the probability of throwing one 6.
And what is t
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Question 326869: What is the probability of drawing one 5 from a pack of cards if you are dealt 4 cards? And
If 3 dice are rolled what is the probability of throwing one 6.
And what is the probability of throwing two 6's.
I would be very very grateful to see your working out for these questions
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Since there are four 5s in the deck of 52 the probability of getting a 5 dealt to you on one draw is 
.
The probability of 
successes in 
trials where 
is the probability of success on any given trial is given by:
\ =\ \left(n\cr k\right\)p^k\left(1\,-\,p\right)^{n\,-\,k})
Where )
is the number of combinations of things taken at a time and is calculated by !})
You want exactly 1 success out of 4 trials where the probability of success on one trial is 
\ =\ \left(4\cr 1\right\)\left(\frac{1}{13}\right)^1\left(\frac{12}{13}\right)^{3})
You can do your own arithmetic. Hint \ =\ n)
.
By the way, note that this is the probability of getting exactly one 5, no more, no less. If you want to know the probability of at least one 5, that is another thing altogether. In that case you could calculate the cumulative probability, that is the probability of one 5 plus the probability of two 5s, and so on:
\ =\ \sum_{i\,=\,1}^4\ \left(4\cr i\right\)\left(\frac{1}{13}\right)^i\left(\frac{12}{13}\right)^{4-i})
Or you could realize that "at least 1" means "any outcome except 0" and then calculate the probability of getting no 5s at all and then subtracting that number from 1.
\ =\ \left(4\cr 0\right\)\left(\frac{1}{13}\right)^0\left(\frac{12}{13}\right)^{4})
Hint: \ =\ 1)
and anything raised to the zero power is 1.
Your other problem is done the same way. You just need to calculate )
and )
John
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