do №%₂ 6 should Cards r) find the 1) probability PPP) None адачу. are drawn from probability of getting atleast of getting getting only P(x=0) Pl.900 pack of 52 con ds 3 diamonds diamond P(x²/3) = 1 - P(x <3) dramonds p(x-4)
do №%₂ 6 should Cards r) find the 1) probability PPP) None адачу. are drawn from probability of getting atleast of getting getting only P(x=0) Pl.900 pack of 52 con ds 3 diamonds diamond P(x²/3) = 1 - P(x <3) dramonds p(x-4)
Algebra and Trigonometry (MindTap Course List)
4th Edition
ISBN:9781305071742
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter14: Counting And Probability
Section14.2: Probability
Problem 3E: The conditional probability of E given that F occurs is P(EF)=___________. So in rolling a die the...
Related questions
Question
![D%
"")
find
the
1) probability
PPT) None
P(AOA)
P(ADBOC) =P (A) +p (B)
PLA00
14
should do again.
are
drawn from
Cards
pack of 52 con ds
probability of getting atleast 3 dramonds
P(x²/3) = 1 - P(x <3)
of getting only
only H
diamond
P(x=0)
P(X=1) = P(x) = n
n(s) = 52
soft
probability
p.^ qn-x
totoro and
success => P(E) & D(E)
n(s)
·3/26
= 23
n=6 f
ung
26
=
6 C₁₂ (31/26) (@23/26)
⇒ >
P(x 3) = 1-P(x<3)
1 = [(p (x²0) + p(x ²1) +P[ X=²)]
¹¯ [*o (³66) * (23)³] + 6q, (³/426) (22)5 +6₂
+ 6 (³26) (2)³ + (()*T
Prof.
Randong (1
⇒ 1 - 1.8035975
diamonds
*
probability of getting only
2
23
P(X= 9) =
ocy (26) (26) ..
timp)
500
F(**
11). None is diamond p(x2x)=P(x26)
0- 47 92.
(1/2)* (2) *
26
11)
P(€) = 6/5₂
всо
of getting
де 1-р
1-3
26
пса
P(x) = "(₂ p* qu-x
ра
1) Atleast 3 diamonds.
SC
осп
n(t) = 6.
N
P(AUB)
dramonds p(x=y)
(2
P
PH
"/") (!)
Valla](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8da4a599-d43a-4c44-9697-4b3fa153f840%2F1f02819b-2cc2-47cd-af7d-159213096f85%2Fqbxawli_processed.jpeg&w=3840&q=75)
Transcribed Image Text:D%
"")
find
the
1) probability
PPT) None
P(AOA)
P(ADBOC) =P (A) +p (B)
PLA00
14
should do again.
are
drawn from
Cards
pack of 52 con ds
probability of getting atleast 3 dramonds
P(x²/3) = 1 - P(x <3)
of getting only
only H
diamond
P(x=0)
P(X=1) = P(x) = n
n(s) = 52
soft
probability
p.^ qn-x
totoro and
success => P(E) & D(E)
n(s)
·3/26
= 23
n=6 f
ung
26
=
6 C₁₂ (31/26) (@23/26)
⇒ >
P(x 3) = 1-P(x<3)
1 = [(p (x²0) + p(x ²1) +P[ X=²)]
¹¯ [*o (³66) * (23)³] + 6q, (³/426) (22)5 +6₂
+ 6 (³26) (2)³ + (()*T
Prof.
Randong (1
⇒ 1 - 1.8035975
diamonds
*
probability of getting only
2
23
P(X= 9) =
ocy (26) (26) ..
timp)
500
F(**
11). None is diamond p(x2x)=P(x26)
0- 47 92.
(1/2)* (2) *
26
11)
P(€) = 6/5₂
всо
of getting
де 1-р
1-3
26
пса
P(x) = "(₂ p* qu-x
ра
1) Atleast 3 diamonds.
SC
осп
n(t) = 6.
N
P(AUB)
dramonds p(x=y)
(2
P
PH
"/") (!)
Valla
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