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antiseptic1488 [7]
3 years ago
12

A coin is tossed 6 times. Find the probability P(of exactly 4 occurrences of heads).

Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
7 0
Let's draw!

H                             1                              T
HT                           2                           HT
HTHT                      3                      HTHT
HTHTHT                 4                  HTHTHT
HTHTHTHT            5             HTHTHTHT
HTHTHTHTHT       6        HTHTHTHTHT

You can count the probabilities using this.

HHHHTT
HHHTHT
HHTHHT
HHTHTH
HHHTTH
HTHHHT
HTTHHH
HTHTHH
HTHHTH

THHHHT
THHHTH
THHTHH
THTHHH
TTHHHH

Therefore, I think the probability is 14/64. Not sure so check my work.
14 combinations
64 outcomes (2+4+8+16+32+64 or 2^n)

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There is a 20% sale on in Topshop. The bag I want is now £60. What was the original cost of my bag?
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How do you solve this one 10x^2-25=x^2
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4 0
3 years ago
Algebra 2A Help !<br><br> Tadakimasu ~
Ludmilka [50]
Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)

first factor into (x-r1)(x-r2)... form

p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)

multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2 

so

p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the  root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real

baseically

(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the  root 1 and has multiplity 1

7 0
3 years ago
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