Answer: 32, -64, 128, -256
multiply by 8 on both sides
(20)8 - 8(4b/8) = (18)8
160 - 4b = 144
subtract 160 on both sides
160 - 160 - 4b = 144 - 160
-4b = -16
divide by -4 on both sides
-4b/-4 = -16/-4
b = 4
The probability the man will win will be 13.23%. And the probability of winning if he wins by getting at least four heads in five flips will be 36.01%.
<h3>How to find that a given condition can be modeled by binomial distribution?</h3>
Binomial distributions consist of n independent Bernoulli trials.
Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))
P(X = x) = ⁿCₓ pˣ (1 - p)⁽ⁿ⁻ˣ⁾
A man wins in a gambling game if he gets two heads in five flips of a biased coin. the probability of getting a head with the coin is 0.7.
Then we have
p = 0.7
n = 5
Then the probability the man will win will be
P(X = 2) = ⁵C₂ (0.7)² (1 - 0.7)⁽⁵⁻²⁾
P(X = 2) = 10 x 0.49 x 0.027
P(X = 2) = 0.1323
P(X = 2) = 13.23%
Then the probability of winning if he wins by getting at least four heads in five flips will be
P(X = 4) = ⁵C₄ (0.7)⁴ (1 - 0.7)⁽⁵⁻⁴⁾
P(X = 4) = 5 x 0.2401 x 0.3
P(X = 4) = 0.3601
P(X = 4) = 36.01%
Learn more about binomial distribution here:
brainly.com/question/13609688
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Answer:
\frac{-177}{286}
Explanation:
Given,
\left(\frac{3}{11}\times\frac{5}{6}\right)-\left(\frac{9}{12}\times \frac{4}{3}\right)+\left(\frac{5}{13}\times \frac{6}{15}\right)
ft(\frac{5}{22}\right )-\left(\frac{36}{36}\right)+\left(\frac{2}{13}\right)
{5}{22}\right )-1+\left(\frac{2}{13}\right)
{5\times 13-286+2\times 22}{286}
{65-286+44}{286}
{109-286}{286}
{-177}{286}
