Calculating the Probability of K=3 When Flipping a Coin 6 Times

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flipping a coin 6 times and let K be the number of crowns that we have after flipping the coin 6 times. The coin has a probability of 0.3 to get the crown.

P(K=3)=?

if we flip the coin 6 times we will have 31 crowns but how i can estimate this series of crowns?
 
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To achieve three here's what you do:

the easy way is to calaulate when we don't achieve three crowns and then subtract 1 from it.

We might not have three crowns if:
1. there was none. which is 0.7^6
2. there was just one crown which is: {6\choose 1} *0.3*0.7^5
3. there were only two crowns which is {6\choose 2} *0.3^2 * 0.7^4
 
i will sum this? how i can estimate 2 and 3 with nCk ?
 
i did it but its says that the result its not correct ...
 
ParisSpart said:
if we flip the coin 6 times we will have 31 crowns but how i can estimate this series of crowns?

I think you mean "...we will have 3 crowns...".

You will have 3 crowns after flipping a coin 6 times in (6 choose 3) different ways, right? And you get a crown with probability p = 0.3, which implies that you will not have a crown with probability q = 1-p = 0.7. Suppose one of the experiments results as HHTTTH (which is one of the (6 choose 3) possible ways to get 3 crowns). The probability of getting this result of the experiment is

0.3^3\times 0.7^3

You have to sum these probabilities for all (6 choose 3) ways.

Your question is just an example for the binomial distribution.
 
its not correct...
 
MathematicalPhysicist said:
To achieve three here's what you do:

the easy way is to calaulate when we don't achieve three crowns and then subtract 1 from it.

We might not have three crowns if:
1. there was none. which is 0.7^6
2. there was just one crown which is: {6\choose 1} *0.3*0.7^5
3. there were only two crowns which is {6\choose 2} *0.3^2 * 0.7^4

Your answer finds P(K>=3), but OP asks for P(K=3).
 
ParisSpart said:
its not correct...

Can you give your reasoning? May I ask what your computational result is?
 
i found it finaly thanks for help man.
 
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