Why did my teacher come to this solution , (probability)?












1














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In a box there are 10 balls with number from 1 to 10.If randomly 6 balls are withdrawn . What is the probability that among the withdrawn balls we get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)



$P=frac{binom{8}{4}}{binom{10}{6}} =frac{70}{210}=0.3333$










share|cite|improve this question



























    1














    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    In a box there are 10 balls with number from 1 to 10.If randomly 6 balls are withdrawn . What is the probability that among the withdrawn balls we get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)



    $P=frac{binom{8}{4}}{binom{10}{6}} =frac{70}{210}=0.3333$










    share|cite|improve this question

























      1












      1








      1







      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      In a box there are 10 balls with number from 1 to 10.If randomly 6 balls are withdrawn . What is the probability that among the withdrawn balls we get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)



      $P=frac{binom{8}{4}}{binom{10}{6}} =frac{70}{210}=0.3333$










      share|cite|improve this question













      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      In a box there are 10 balls with number from 1 to 10.If randomly 6 balls are withdrawn . What is the probability that among the withdrawn balls we get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






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      share|cite|improve this question











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      asked Dec 5 '18 at 11:06









      Johnny AdamsJohnny Adams

      227




      227






















          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.






          share|cite|improve this answer





















          • I guess I need a better understanding of combinations
            – Johnny Adams
            Dec 5 '18 at 13:39











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          3














          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.






          share|cite|improve this answer





















          • I guess I need a better understanding of combinations
            – Johnny Adams
            Dec 5 '18 at 13:39
















          3














          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.






          share|cite|improve this answer





















          • I guess I need a better understanding of combinations
            – Johnny Adams
            Dec 5 '18 at 13:39














          3












          3








          3






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • 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


















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