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True [87]
3 years ago
10

Five independent flips of a fair coin are made. Find the probability that(a) the first three flips are the same;(b) either the f

irst three flips are the same, or the last three flips are the same;(c) there are at least two heads among the first three flips, and at least two tails amongthe last three flips
Mathematics
1 answer:
Korvikt [17]3 years ago
5 0

Answer:

The answers to the question are

(a) 1/4

(b) 7/16

(c) 1/16

Step-by-step explanation:

To solve the question we note that

Total number  of outcomes = 32

Probability of the event of first three flips are the same =P(F)

Probability of the event of last three flips are the same =P(L)

Total number of outcomes = 2⁵ = 32

The number of ways in which the frist three flips are the same is

F  TTTTT, TTTTH, TTTHH, HHHTT, HHHHT, HHHHH, TTTHT, HHHTH = 8

L:  TTTTT, HHTTT, HTTTT, THTTT, HHHHH, TTHHH, HTHHH, THHHH = 8

The probability that the first and the last three flips are the same that is

F ∩ L; TTTTT, HHHHH = 2

Therefore P(F ∩ L) = 2/32

(a) P(F) = 8/32 =1/4 also

     P(L) = 8/32 =1/4

(b) P(LUF) =  P(L) + P(F) - P(F ∩ L) = 1/4+1/4-1/16 =7/16

(c) Let the event of at least two heads among the first three flips be H

and the event of at least two tails among the last three flips be T

Then we have

H; HHHHH, HHHHT, HHHTT, HHHTH, HHTTT, HHTHH, THHTT, THHHT

= 8

T; TTTTT. HTTTT, HHTTT, THTTT, THHTT, HHHTT, THHTT, HTTTH =8

Also H∩T  = TTTHH, HHTTT = 2

Therefore P(H ∩ T) = 2/32 = 1/16

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<h3>The missing value in the given solution is x=6</h3><h3>Therefore the solution is (6,-4)</h3>

Step-by-step explanation:

Given equation is 6x+7y=4x+4y\hfill (1)

<h3>To find the missing value in the solution to the equation :</h3>
  • Let missing value in the solution be x
  • Then the solution of the given equation is (x,-4)

Substitute the value of y=-4 in the given equation (1) we get

6x+7(-4)=4x+4(-4)

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6x-28-(4x-16)=4x-16-(4x-16)

6x-28-4x+16=4x-16-4x+16 ( adding the like terms )

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10.) state whether the given side lengths can form a triangle
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Answer:

A. 5,8,12, side lengths can form a triangle.

B. 20, 18, 2, side lengths can not form a triangle.

C. 15, 26 , 9, side lengths can not form a triangle.

D. 4.75, 12.25, 16.25, side lengths can form a triangle.

Step-by-step explanation:

We are given three sides in options A, B , C and D.

We need to check if the given sides would form a triangle or not.

Note: Sum of two sides is always greater than third sides in a triangle.

So, we need to check the sum of two sides of given sides length to check if sum is greater than third side length.

A. 5,8,12.

5+8 is greater than 12.

5+12 is greater than 8.

12+8 is greater than 5.

Therefore A. 5,8,12, side lengths can form a triangle.

B. 20, 18, 2

18+2 is not greater than 20.

Therefore B. 20, 18, 2, side lengths can not form a triangle.

C. 15, 26 , 9

15+9 is less than 26.

Therefore C. 15, 26 , 9, side lengths can not form a triangle.

D. 4.75, 12.25, 16.25.

4.75+12.25 is greater than 16.25.

4.75+16.25 is greater than 12.25.

12.25+16.25 is greater than 4.75.

Therefore D. 4.75, 12.25, 16.25, side lengths can form a triangle.


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