Question
ok we got this packet that has an answer key and we g0tta show our work of how to get the answer. okayyy... they all seem like really easy problems s0 i think im just dumbbb
"a bag has 6 red and 2 white marbles. If two marbles are selected, what is the probability that one is red and the other is white?" - the answer is 3/7. how do i get that??
it seems like a real easy problem, i know. but i cant seem to get it :|
next. "Jose has 5 dimes, 7 nickels, and 4 pennies in his pocket. he selects 4 coins. What is the probability that he has 2 dimes and 2 pennies." -answer is 3/91.
Last one. "If a card is selected from a deck of cards, find the probability that it is not red or not a face card." answer is 23/26. This one really doesnt make sense ???
"a bag has 6 red and 2 white marbles. If two marbles are selected, what is the probability that one is red and the other is white?" - the answer is 3/7. how do i get that??
it seems like a real easy problem, i know. but i cant seem to get it :|
next. "Jose has 5 dimes, 7 nickels, and 4 pennies in his pocket. he selects 4 coins. What is the probability that he has 2 dimes and 2 pennies." -answer is 3/91.
Last one. "If a card is selected from a deck of cards, find the probability that it is not red or not a face card." answer is 23/26. This one really doesnt make sense ???
Answers
Reiny
first one:
C(6,1)*C(2,1)/C8,2) = 6*2/28 = 3/7
second one:
C(5,2)*C(4,2)/C(16,4) = 10*6/1820 = 3/91
third one:
the key word is the "or"
prob(not red OR not a facecard)
= Prob(not red) + prob(not a facecard) - prob(not red AND not a facecard)
= 26/52 + 40/52 - 20/52
= 46/52
= 23/28
C(6,1)*C(2,1)/C8,2) = 6*2/28 = 3/7
second one:
C(5,2)*C(4,2)/C(16,4) = 10*6/1820 = 3/91
third one:
the key word is the "or"
prob(not red OR not a facecard)
= Prob(not red) + prob(not a facecard) - prob(not red AND not a facecard)
= 26/52 + 40/52 - 20/52
= 46/52
= 23/28
John
thank you so much!