Probability of Picking k Winning Numbers in a Lottery

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SUMMARY

The discussion focuses on calculating the probability of selecting k winning numbers in a lottery where 6 distinct numbers are drawn from a set of 50. The initial calculations presented for k=0 to k=6 were incorrect as they did not account for the combinations of winning numbers. The correct formula for k winning numbers is given by the combination of selecting k winners from 6 (6Ck) multiplied by the combination of selecting the remaining numbers from the 44 non-winning numbers (44C(6-k)). This adjustment ensures that the probabilities sum to 1 across all values of k.

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Punkyc7
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In a lottery , 6 distinct numbers are selected at random from a set {1...50} and designated as a winning number. The player picks six numbers beforehand hoping to include as many numbers as possible. Find the probability that the user picks k winning numbers for k=1,2..,6
So my one concern is that my answers don't sum to 1for K=0 I have (44C6)/(50C6)
for K=1 I have (44C5)/(50C6)
for K=2 I have (44C4)/(50C6)
for K=3 I have (44C3)/(50C6)
for K=4 I have (44C2)/(50C6)
for K=5 I have (44C1)/(50C6)
for K=6 I have (44C0)/(50C6)

My reasoning is that for k=0 you didnt pick any of the six numbers so you must have chosen 6 of the 44 remaining
and used similar reasoning for k=1...6
My question is are my answers right or did i miss something ?
 
Last edited:
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Let's consider the case K=2, for example. Let's call your chosen 6 numbers "winners" and the remaining 44 numbers "losers". 44C4 is the number of ways you can pick the 4 losers-- there you are right.

But how many ways are there for you to pick the 2 winners?
 
There are 6C2. So there should be 6C2 * 44C4Thanks that fixed it
 
Last edited:

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