MATH: statistic/ probalitity - need help problem with #9-19 (a-c, a-b) - related
ID: 3065636 • Letter: M
Question
MATH: statistic/ probalitity
- need help problem with #9-19 (a-c, a-b)
- related questions
- please explain/show steps/answer to the question
Explanation / Answer
There are 52 cards, one card can be selected in 52C1.
a)
P(King) = 4C1/52C1 = 4/52 = 1/13
P(Queen) = 4C1/52C1 = 4/52 = 1/13
P(club) = 13C1/52C1 = 1/4
b)
There are 16 (13 + 4 - 1) cards which are king or club.
P(king or club) = 16C1/52C1 = 16/52 = 4/13
from part a)
P(King) + P(club) - P(King and Club)
= 1/13 + 1/4 - 1/52 = 4/13
c)
There are 8 cards (4+4) which are king or queen.
P(king or queen) = 8/52 = 2/13
From part a) P(king) + P(queen) = 1/13 + 1/13 = 2/13
9-19
a)
There are 4*4 = 16 cards which are less than 5
P(less than 5) = 16C1/52C1 = 16/52 = 4/13
b)
26 red cards
12 face cards
6 face cards which are red
Required probability = (26 + 12 - 6)/52 = 32/52 = 8/13
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