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Anita505
Group Title
A card is drawn at random from a standard 52card deck. Events G and H are:
G= the drawn card is black
H= the drawn card is divisible by 5 (face cards are not valued)
(A)Find P(HIG)
(B)Test H and G for independence
(A)P(HIG)=
 one year ago
 one year ago
Anita505 Group Title
A card is drawn at random from a standard 52card deck. Events G and H are: G= the drawn card is black H= the drawn card is divisible by 5 (face cards are not valued) (A)Find P(HIG) (B)Test H and G for independence (A)P(HIG)=
 one year ago
 one year ago

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kropot72 Group TitleBest ResponseYou've already chosen the best response.0
P(HG) means the probability of H given G. Given that the drawn card is black, the sample space is 26 cards. There are 4 black cards that are either a 5 or a 10. \[P(HG)=\frac{4}{26}=\frac{2}{13}\]
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
@Anita505 Are you there?
 one year ago

Anita505 Group TitleBest ResponseYou've already chosen the best response.0
yes i am here :)
 one year ago

Anita505 Group TitleBest ResponseYou've already chosen the best response.0
Thank you for the help! @kropot72
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
You're welcome :)
 one year ago

Anita505 Group TitleBest ResponseYou've already chosen the best response.0
however would Test H and G for independence... would it be independent? or dependent?
 one year ago

Anita505 Group TitleBest ResponseYou've already chosen the best response.0
oh its independent
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
If two events A and B are independent, then the probability of A occurring is unaffected by whether of not B has occurred. Therefore if the events are independent P(AB) = P(A) In this case we test whether P(HG) = P(H) There are 8 cards that are 5 or 10 and either black or red. Therefore \[P(H)=\frac{8}{52}=\frac{2}{13}\] Have we confirmed that H and G are independent?
 one year ago

Anita505 Group TitleBest ResponseYou've already chosen the best response.0
Yes this is independent :) thank you once again for your help! much appreciated!
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
Correct! You're welcome :)
 one year ago
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