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4 years ago

Please help me on this

Mathematics

1 answer:

Airida [17]4 years ago

4 0

First one go 5 up
second one go 18 up
third one go 20 up
fourth one go 17 up
fifth one go 16 up
sixth one go 23 up
seventh one go 19 up
tenth one go 14 up
eleventh one go 8 up
twelfth one go 8 up
thirteenth one go 6 up
fourteenth one go 12 up
fiftheenth one go 20 up
sixteenth one go 14 up
seventeenth one go 18 up

this was the best I could explain

use the numbers as the question shows
I hope I helped

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41% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a pr

Amiraneli [1.4K]

Answer:

a) $P(X=5)=0.208$

b) $P(X\geq 6)=0.1854$

c) $P(X$

Step-by-step explanation:

Given : 41% of men consider themselves professional baseball fans. You randomly select 10 men and ask each if he considers himself a professional baseball fan.

To Find : The probability that the number who consider themselves baseball fans is (a) exactly five, (b) at least six, and (c) less than four.

Solution :

Applying binomial theorem,

$P(X=r)=^nC_r\times p^r\times (1-p)^{n-r}$

The success p= 41%=0.41

The failure (1-p)=1-041=0.59

Number of selection n=10

a) The probability that the number who consider themselves baseball fans is exactly five,

i.e. X=5

$P(X=5)=^{10}C_5\times (0.41)^5\times (0.59)^{10-5}$

$P(X=5)=\frac{10!}{5!5!}\times (0.41)^5\times (0.59)^{5}$

$P(X=5)=252\times 0.0115\times 0.071$

$P(X=5)=0.208$

b) The probability that the number who consider themselves baseball fans is at least six,

i.e. $X\geq 6$

$P(X\geq 6)=1-P(X$

$P(X\geq 6)=1-[P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)]$

$P(X\geq 6)=1-[^{10}C_0\times (0.41)^0\times (0.59)^{10-0}+^{10}C_1\times (0.41)^1\times (0.59)^{10-1}+^{10}C_2\times (0.41)^2\times (0.59)^{10-2}+^{10}C_3\times (0.41)^3\times (0.59)^{10-3}+^{10}C_4\times (0.41)^4\times (0.59)^{10-4}+^{10}C_5\times (0.41)^5\times (0.59)^{10-5}]$