What is the probability of a card of heart or non jack drawn from a standard deck?

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According to the formula UNION of $2$ Events: $$P(Ecup F)=P(E)+P(F)−P(Ecap F)
- P(textheart) = frac1352, P(textnon-jack) = frac4852, P(textheart cap textnon-jack)= frac1252\ implies P(textheart cup textnon-jack) = frac4952.$$
But this result does not make sense. I think the answer must be $48$, because there is none of any of the $4$ jacks drawn.
So I would like to know your own idea about this, please? Thank you all!




probability




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  • 1




    So the allowed cards are all the hearts, and all non-jacks? That is, all the cards which are not jacks, and the jack of hearts? Because those are indeed 49 card.s
    – CompuChip
    Aug 29 at 10:45














up vote
0
down vote

favorite












According to the formula UNION of $2$ Events: $$P(Ecup F)=P(E)+P(F)−P(Ecap F)
- P(textheart) = frac1352, P(textnon-jack) = frac4852, P(textheart cap textnon-jack)= frac1252\ implies P(textheart cup textnon-jack) = frac4952.$$
But this result does not make sense. I think the answer must be $48$, because there is none of any of the $4$ jacks drawn.
So I would like to know your own idea about this, please? Thank you all!




probability




share|cite|improve this question


















  • 1




    So the allowed cards are all the hearts, and all non-jacks? That is, all the cards which are not jacks, and the jack of hearts? Because those are indeed 49 card.s
    – CompuChip
    Aug 29 at 10:45












up vote
0
down vote

favorite









up vote
0
down vote

favorite











According to the formula UNION of $2$ Events: $$P(Ecup F)=P(E)+P(F)−P(Ecap F)
- P(textheart) = frac1352, P(textnon-jack) = frac4852, P(textheart cap textnon-jack)= frac1252\ implies P(textheart cup textnon-jack) = frac4952.$$
But this result does not make sense. I think the answer must be $48$, because there is none of any of the $4$ jacks drawn.
So I would like to know your own idea about this, please? Thank you all!




probability




share|cite|improve this question














According to the formula UNION of $2$ Events: $$P(Ecup F)=P(E)+P(F)−P(Ecap F)
- P(textheart) = frac1352, P(textnon-jack) = frac4852, P(textheart cap textnon-jack)= frac1252\ implies P(textheart cup textnon-jack) = frac4952.$$
But this result does not make sense. I think the answer must be $48$, because there is none of any of the $4$ jacks drawn.
So I would like to know your own idea about this, please? Thank you all!





probability





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edited Aug 29 at 11:13









amWhy

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asked Aug 29 at 10:36









Willy

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122







  • 1




    So the allowed cards are all the hearts, and all non-jacks? That is, all the cards which are not jacks, and the jack of hearts? Because those are indeed 49 card.s
    – CompuChip
    Aug 29 at 10:45












  • 1




    So the allowed cards are all the hearts, and all non-jacks? That is, all the cards which are not jacks, and the jack of hearts? Because those are indeed 49 card.s
    – CompuChip
    Aug 29 at 10:45







1




1




So the allowed cards are all the hearts, and all non-jacks? That is, all the cards which are not jacks, and the jack of hearts? Because those are indeed 49 card.s
– CompuChip
Aug 29 at 10:45




So the allowed cards are all the hearts, and all non-jacks? That is, all the cards which are not jacks, and the jack of hearts? Because those are indeed 49 card.s
– CompuChip
Aug 29 at 10:45










2 Answers
2






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0
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I think my problem is about word. "a heart OR a non jack" means that jack of heart is included as a probability. So 49/52 is the right number.






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  • Yes, this is correct.
    – Tanner Swett
    Aug 30 at 3:30

















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0
down vote













Try looking at the problem this way:
We either draw a jack or not a jack. The probability of not drawing a jack is $frac4852$.
If we do draw a jack, we are interested in the case that it is the jack of hearts. So the probability of drawing a jack and it being the jack of hearts is
$frac14 frac452=frac152$. Now since the events jack and not-a-jack are disjoint we can simply add the probabilities to find the desired probability.



In your terminology: take E=draw a jack and F=Draw a non-jack which is the jack of hearts. Clearly $P(Ecap F)=0$.






share|cite|improve this answer






















  • Thanks Jan! But the probability of drawing a jack 48/52 is a mistake, should be 4/52.
    – Willy
    Aug 29 at 23:44










  • Ah I see what you mean, it should say of not drawing a jack. My apologies I will edit it.
    – Jan
    Aug 30 at 6:51










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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













I think my problem is about word. "a heart OR a non jack" means that jack of heart is included as a probability. So 49/52 is the right number.






share|cite|improve this answer




















  • Yes, this is correct.
    – Tanner Swett
    Aug 30 at 3:30














up vote
0
down vote













I think my problem is about word. "a heart OR a non jack" means that jack of heart is included as a probability. So 49/52 is the right number.






share|cite|improve this answer




















  • Yes, this is correct.
    – Tanner Swett
    Aug 30 at 3:30












up vote
0
down vote










up vote
0
down vote









I think my problem is about word. "a heart OR a non jack" means that jack of heart is included as a probability. So 49/52 is the right number.






share|cite|improve this answer












I think my problem is about word. "a heart OR a non jack" means that jack of heart is included as a probability. So 49/52 is the right number.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 29 at 23:48









Willy

122




122











  • Yes, this is correct.
    – Tanner Swett
    Aug 30 at 3:30
















  • Yes, this is correct.
    – Tanner Swett
    Aug 30 at 3:30















Yes, this is correct.
– Tanner Swett
Aug 30 at 3:30




Yes, this is correct.
– Tanner Swett
Aug 30 at 3:30










up vote
0
down vote













Try looking at the problem this way:
We either draw a jack or not a jack. The probability of not drawing a jack is $frac4852$.
If we do draw a jack, we are interested in the case that it is the jack of hearts. So the probability of drawing a jack and it being the jack of hearts is
$frac14 frac452=frac152$. Now since the events jack and not-a-jack are disjoint we can simply add the probabilities to find the desired probability.



In your terminology: take E=draw a jack and F=Draw a non-jack which is the jack of hearts. Clearly $P(Ecap F)=0$.






share|cite|improve this answer






















  • Thanks Jan! But the probability of drawing a jack 48/52 is a mistake, should be 4/52.
    – Willy
    Aug 29 at 23:44










  • Ah I see what you mean, it should say of not drawing a jack. My apologies I will edit it.
    – Jan
    Aug 30 at 6:51














up vote
0
down vote













Try looking at the problem this way:
We either draw a jack or not a jack. The probability of not drawing a jack is $frac4852$.
If we do draw a jack, we are interested in the case that it is the jack of hearts. So the probability of drawing a jack and it being the jack of hearts is
$frac14 frac452=frac152$. Now since the events jack and not-a-jack are disjoint we can simply add the probabilities to find the desired probability.



In your terminology: take E=draw a jack and F=Draw a non-jack which is the jack of hearts. Clearly $P(Ecap F)=0$.






share|cite|improve this answer






















  • Thanks Jan! But the probability of drawing a jack 48/52 is a mistake, should be 4/52.
    – Willy
    Aug 29 at 23:44










  • Ah I see what you mean, it should say of not drawing a jack. My apologies I will edit it.
    – Jan
    Aug 30 at 6:51












up vote
0
down vote










up vote
0
down vote









Try looking at the problem this way:
We either draw a jack or not a jack. The probability of not drawing a jack is $frac4852$.
If we do draw a jack, we are interested in the case that it is the jack of hearts. So the probability of drawing a jack and it being the jack of hearts is
$frac14 frac452=frac152$. Now since the events jack and not-a-jack are disjoint we can simply add the probabilities to find the desired probability.



In your terminology: take E=draw a jack and F=Draw a non-jack which is the jack of hearts. Clearly $P(Ecap F)=0$.






share|cite|improve this answer














Try looking at the problem this way:
We either draw a jack or not a jack. The probability of not drawing a jack is $frac4852$.
If we do draw a jack, we are interested in the case that it is the jack of hearts. So the probability of drawing a jack and it being the jack of hearts is
$frac14 frac452=frac152$. Now since the events jack and not-a-jack are disjoint we can simply add the probabilities to find the desired probability.



In your terminology: take E=draw a jack and F=Draw a non-jack which is the jack of hearts. Clearly $P(Ecap F)=0$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 30 at 6:52

























answered Aug 29 at 11:03









Jan

701417




701417











  • Thanks Jan! But the probability of drawing a jack 48/52 is a mistake, should be 4/52.
    – Willy
    Aug 29 at 23:44










  • Ah I see what you mean, it should say of not drawing a jack. My apologies I will edit it.
    – Jan
    Aug 30 at 6:51
















  • Thanks Jan! But the probability of drawing a jack 48/52 is a mistake, should be 4/52.
    – Willy
    Aug 29 at 23:44










  • Ah I see what you mean, it should say of not drawing a jack. My apologies I will edit it.
    – Jan
    Aug 30 at 6:51















Thanks Jan! But the probability of drawing a jack 48/52 is a mistake, should be 4/52.
– Willy
Aug 29 at 23:44




Thanks Jan! But the probability of drawing a jack 48/52 is a mistake, should be 4/52.
– Willy
Aug 29 at 23:44












Ah I see what you mean, it should say of not drawing a jack. My apologies I will edit it.
– Jan
Aug 30 at 6:51




Ah I see what you mean, it should say of not drawing a jack. My apologies I will edit it.
– Jan
Aug 30 at 6:51

















 

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