Why did my teacher come to this solution , (probability)?
Here is the problem :
In a box there are 10 balls with number from 1 to 10.If randomly 6 balls are withdrawn . What is the probability that among the withdrawn balls we get
a) the ball with number 1 and b) the balls with numbers 1 and 2 .
For a) she solved like this
$P=frac{binom{9}{5}}{binom{10}{6}} =frac{126}{210}=0.6$
and for b)
$P=frac{binom{8}{4}}{binom{10}{6}} =frac{70}{210}=0.3333$
probability
add a comment |
Here is the problem :
In a box there are 10 balls with number from 1 to 10.If randomly 6 balls are withdrawn . What is the probability that among the withdrawn balls we get
a) the ball with number 1 and b) the balls with numbers 1 and 2 .
For a) she solved like this
$P=frac{binom{9}{5}}{binom{10}{6}} =frac{126}{210}=0.6$
and for b)
$P=frac{binom{8}{4}}{binom{10}{6}} =frac{70}{210}=0.3333$
probability
add a comment |
Here is the problem :
In a box there are 10 balls with number from 1 to 10.If randomly 6 balls are withdrawn . What is the probability that among the withdrawn balls we get
a) the ball with number 1 and b) the balls with numbers 1 and 2 .
For a) she solved like this
$P=frac{binom{9}{5}}{binom{10}{6}} =frac{126}{210}=0.6$
and for b)
$P=frac{binom{8}{4}}{binom{10}{6}} =frac{70}{210}=0.3333$
probability
Here is the problem :
In a box there are 10 balls with number from 1 to 10.If randomly 6 balls are withdrawn . What is the probability that among the withdrawn balls we get
a) the ball with number 1 and b) the balls with numbers 1 and 2 .
For a) she solved like this
$P=frac{binom{9}{5}}{binom{10}{6}} =frac{126}{210}=0.6$
and for b)
$P=frac{binom{8}{4}}{binom{10}{6}} =frac{70}{210}=0.3333$
probability
probability
asked Dec 5 '18 at 11:06
Johnny AdamsJohnny Adams
227
227
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1 Answer
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Since we are only interested in the numbers of the balls, the order does not matter. Consider the event that the ball with number $1$ is withdrawn among the $6$ balls. That means that one ball is fixed (this is the ball with number $1$) and $5$ balls can be chosen freely. There are ${9choose 5}$ ways to choose $5$ balls out of $9$ balls when the order is not important. We need to divide ${9choose 5}$ by the number of ways to choose $6$ balls out of $10$ when the order is not important.
I hope this helps.
I guess I need a better understanding of combinations
– Johnny Adams
Dec 5 '18 at 13:39
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
Since we are only interested in the numbers of the balls, the order does not matter. Consider the event that the ball with number $1$ is withdrawn among the $6$ balls. That means that one ball is fixed (this is the ball with number $1$) and $5$ balls can be chosen freely. There are ${9choose 5}$ ways to choose $5$ balls out of $9$ balls when the order is not important. We need to divide ${9choose 5}$ by the number of ways to choose $6$ balls out of $10$ when the order is not important.
I hope this helps.
I guess I need a better understanding of combinations
– Johnny Adams
Dec 5 '18 at 13:39
add a comment |
Since we are only interested in the numbers of the balls, the order does not matter. Consider the event that the ball with number $1$ is withdrawn among the $6$ balls. That means that one ball is fixed (this is the ball with number $1$) and $5$ balls can be chosen freely. There are ${9choose 5}$ ways to choose $5$ balls out of $9$ balls when the order is not important. We need to divide ${9choose 5}$ by the number of ways to choose $6$ balls out of $10$ when the order is not important.
I hope this helps.
I guess I need a better understanding of combinations
– Johnny Adams
Dec 5 '18 at 13:39
add a comment |
Since we are only interested in the numbers of the balls, the order does not matter. Consider the event that the ball with number $1$ is withdrawn among the $6$ balls. That means that one ball is fixed (this is the ball with number $1$) and $5$ balls can be chosen freely. There are ${9choose 5}$ ways to choose $5$ balls out of $9$ balls when the order is not important. We need to divide ${9choose 5}$ by the number of ways to choose $6$ balls out of $10$ when the order is not important.
I hope this helps.
Since we are only interested in the numbers of the balls, the order does not matter. Consider the event that the ball with number $1$ is withdrawn among the $6$ balls. That means that one ball is fixed (this is the ball with number $1$) and $5$ balls can be chosen freely. There are ${9choose 5}$ ways to choose $5$ balls out of $9$ balls when the order is not important. We need to divide ${9choose 5}$ by the number of ways to choose $6$ balls out of $10$ when the order is not important.
I hope this helps.
answered Dec 5 '18 at 11:12
Cm7F7BbCm7F7Bb
12.4k32142
12.4k32142
I guess I need a better understanding of combinations
– Johnny Adams
Dec 5 '18 at 13:39
add a comment |
I guess I need a better understanding of combinations
– Johnny Adams
Dec 5 '18 at 13:39
I guess I need a better understanding of combinations
– Johnny Adams
Dec 5 '18 at 13:39
I guess I need a better understanding of combinations
– Johnny Adams
Dec 5 '18 at 13:39
add a comment |
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