Probability of drawing same ball until draw 4 or 'greater'?












0












$begingroup$


I'm working on the series of two different questions.
I got the first problem without difficulty but got stuck with second problem.



Here is the first problem I'm writing for context.



You have a bag containing $2$ red balls, $4$ green balls, and $5$ purple balls. You draw one ball from the bag: it's green. Call this draw $0$. Keep this ball and continue drawing (without replacement) until you get another green ball. Call these draws $1, 2...k$. What is the probability that you don't draw another green ball until draw $3$?



I got the answer by $largefrac{7}{10} cdot frac{6}{9} cdot frac{3}{8} = 0.175$, which was right.
However, next problem is somewhat tricky by its wording.



Resetting the original scenario such that draw $0$ yields a green ball, what is the probability that you don't draw another green ball until draw $4$ or greater?



I can't understand how can I get the answer for '$4$ or greater'. I just assumed it similar to first one and ended up with the answer $0.125 (largefrac{7}{10} cdot frac{6}{9} cdot frac{5}{8} cdot frac{3}{7})$ but it wasn't accurate.



What should I do for '$4$ or greater'?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: Getting the first green ball at draw $4$ or later is the same as saying "none of the draws up until draw $4$ are green."
    $endgroup$
    – platty
    Dec 8 '18 at 22:00










  • $begingroup$
    Hi platty, what's the difference between what you just said 'none of the draws up until draw 4 are green' and my last answer (7/10 * 6/9 * 5/8 * 3/7)?
    $endgroup$
    – Daniel Kim
    Dec 8 '18 at 22:08










  • $begingroup$
    never mind, I got it now! thank you for hint!
    $endgroup$
    – Daniel Kim
    Dec 8 '18 at 22:10










  • $begingroup$
    Notice that the event that you don't draw another green ball until draw $4$ or later is the complement of the event that you draw another green ball during the first three draws.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 22:39










  • $begingroup$
    Leave out the $frac{3}{7}$ factor in your answer.
    $endgroup$
    – herb steinberg
    Dec 8 '18 at 22:52
















0












$begingroup$


I'm working on the series of two different questions.
I got the first problem without difficulty but got stuck with second problem.



Here is the first problem I'm writing for context.



You have a bag containing $2$ red balls, $4$ green balls, and $5$ purple balls. You draw one ball from the bag: it's green. Call this draw $0$. Keep this ball and continue drawing (without replacement) until you get another green ball. Call these draws $1, 2...k$. What is the probability that you don't draw another green ball until draw $3$?



I got the answer by $largefrac{7}{10} cdot frac{6}{9} cdot frac{3}{8} = 0.175$, which was right.
However, next problem is somewhat tricky by its wording.



Resetting the original scenario such that draw $0$ yields a green ball, what is the probability that you don't draw another green ball until draw $4$ or greater?



I can't understand how can I get the answer for '$4$ or greater'. I just assumed it similar to first one and ended up with the answer $0.125 (largefrac{7}{10} cdot frac{6}{9} cdot frac{5}{8} cdot frac{3}{7})$ but it wasn't accurate.



What should I do for '$4$ or greater'?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Hint: Getting the first green ball at draw $4$ or later is the same as saying "none of the draws up until draw $4$ are green."
    $endgroup$
    – platty
    Dec 8 '18 at 22:00










  • $begingroup$
    Hi platty, what's the difference between what you just said 'none of the draws up until draw 4 are green' and my last answer (7/10 * 6/9 * 5/8 * 3/7)?
    $endgroup$
    – Daniel Kim
    Dec 8 '18 at 22:08










  • $begingroup$
    never mind, I got it now! thank you for hint!
    $endgroup$
    – Daniel Kim
    Dec 8 '18 at 22:10










  • $begingroup$
    Notice that the event that you don't draw another green ball until draw $4$ or later is the complement of the event that you draw another green ball during the first three draws.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 22:39










  • $begingroup$
    Leave out the $frac{3}{7}$ factor in your answer.
    $endgroup$
    – herb steinberg
    Dec 8 '18 at 22:52














0












0








0





$begingroup$


I'm working on the series of two different questions.
I got the first problem without difficulty but got stuck with second problem.



Here is the first problem I'm writing for context.



You have a bag containing $2$ red balls, $4$ green balls, and $5$ purple balls. You draw one ball from the bag: it's green. Call this draw $0$. Keep this ball and continue drawing (without replacement) until you get another green ball. Call these draws $1, 2...k$. What is the probability that you don't draw another green ball until draw $3$?



I got the answer by $largefrac{7}{10} cdot frac{6}{9} cdot frac{3}{8} = 0.175$, which was right.
However, next problem is somewhat tricky by its wording.



Resetting the original scenario such that draw $0$ yields a green ball, what is the probability that you don't draw another green ball until draw $4$ or greater?



I can't understand how can I get the answer for '$4$ or greater'. I just assumed it similar to first one and ended up with the answer $0.125 (largefrac{7}{10} cdot frac{6}{9} cdot frac{5}{8} cdot frac{3}{7})$ but it wasn't accurate.



What should I do for '$4$ or greater'?










share|cite|improve this question











$endgroup$




I'm working on the series of two different questions.
I got the first problem without difficulty but got stuck with second problem.



Here is the first problem I'm writing for context.



You have a bag containing $2$ red balls, $4$ green balls, and $5$ purple balls. You draw one ball from the bag: it's green. Call this draw $0$. Keep this ball and continue drawing (without replacement) until you get another green ball. Call these draws $1, 2...k$. What is the probability that you don't draw another green ball until draw $3$?



I got the answer by $largefrac{7}{10} cdot frac{6}{9} cdot frac{3}{8} = 0.175$, which was right.
However, next problem is somewhat tricky by its wording.



Resetting the original scenario such that draw $0$ yields a green ball, what is the probability that you don't draw another green ball until draw $4$ or greater?



I can't understand how can I get the answer for '$4$ or greater'. I just assumed it similar to first one and ended up with the answer $0.125 (largefrac{7}{10} cdot frac{6}{9} cdot frac{5}{8} cdot frac{3}{7})$ but it wasn't accurate.



What should I do for '$4$ or greater'?







probability statistics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 22:11









Gaby Alfonso

839316




839316










asked Dec 8 '18 at 21:53









Daniel KimDaniel Kim

111




111








  • 1




    $begingroup$
    Hint: Getting the first green ball at draw $4$ or later is the same as saying "none of the draws up until draw $4$ are green."
    $endgroup$
    – platty
    Dec 8 '18 at 22:00










  • $begingroup$
    Hi platty, what's the difference between what you just said 'none of the draws up until draw 4 are green' and my last answer (7/10 * 6/9 * 5/8 * 3/7)?
    $endgroup$
    – Daniel Kim
    Dec 8 '18 at 22:08










  • $begingroup$
    never mind, I got it now! thank you for hint!
    $endgroup$
    – Daniel Kim
    Dec 8 '18 at 22:10










  • $begingroup$
    Notice that the event that you don't draw another green ball until draw $4$ or later is the complement of the event that you draw another green ball during the first three draws.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 22:39










  • $begingroup$
    Leave out the $frac{3}{7}$ factor in your answer.
    $endgroup$
    – herb steinberg
    Dec 8 '18 at 22:52














  • 1




    $begingroup$
    Hint: Getting the first green ball at draw $4$ or later is the same as saying "none of the draws up until draw $4$ are green."
    $endgroup$
    – platty
    Dec 8 '18 at 22:00










  • $begingroup$
    Hi platty, what's the difference between what you just said 'none of the draws up until draw 4 are green' and my last answer (7/10 * 6/9 * 5/8 * 3/7)?
    $endgroup$
    – Daniel Kim
    Dec 8 '18 at 22:08










  • $begingroup$
    never mind, I got it now! thank you for hint!
    $endgroup$
    – Daniel Kim
    Dec 8 '18 at 22:10










  • $begingroup$
    Notice that the event that you don't draw another green ball until draw $4$ or later is the complement of the event that you draw another green ball during the first three draws.
    $endgroup$
    – N. F. Taussig
    Dec 8 '18 at 22:39










  • $begingroup$
    Leave out the $frac{3}{7}$ factor in your answer.
    $endgroup$
    – herb steinberg
    Dec 8 '18 at 22:52








1




1




$begingroup$
Hint: Getting the first green ball at draw $4$ or later is the same as saying "none of the draws up until draw $4$ are green."
$endgroup$
– platty
Dec 8 '18 at 22:00




$begingroup$
Hint: Getting the first green ball at draw $4$ or later is the same as saying "none of the draws up until draw $4$ are green."
$endgroup$
– platty
Dec 8 '18 at 22:00












$begingroup$
Hi platty, what's the difference between what you just said 'none of the draws up until draw 4 are green' and my last answer (7/10 * 6/9 * 5/8 * 3/7)?
$endgroup$
– Daniel Kim
Dec 8 '18 at 22:08




$begingroup$
Hi platty, what's the difference between what you just said 'none of the draws up until draw 4 are green' and my last answer (7/10 * 6/9 * 5/8 * 3/7)?
$endgroup$
– Daniel Kim
Dec 8 '18 at 22:08












$begingroup$
never mind, I got it now! thank you for hint!
$endgroup$
– Daniel Kim
Dec 8 '18 at 22:10




$begingroup$
never mind, I got it now! thank you for hint!
$endgroup$
– Daniel Kim
Dec 8 '18 at 22:10












$begingroup$
Notice that the event that you don't draw another green ball until draw $4$ or later is the complement of the event that you draw another green ball during the first three draws.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 22:39




$begingroup$
Notice that the event that you don't draw another green ball until draw $4$ or later is the complement of the event that you draw another green ball during the first three draws.
$endgroup$
– N. F. Taussig
Dec 8 '18 at 22:39












$begingroup$
Leave out the $frac{3}{7}$ factor in your answer.
$endgroup$
– herb steinberg
Dec 8 '18 at 22:52




$begingroup$
Leave out the $frac{3}{7}$ factor in your answer.
$endgroup$
– herb steinberg
Dec 8 '18 at 22:52










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