Answer: 0.05
Step-by-step explanation:
Let M = Event of getting an A in Marketing class.
S = Event of getting an A in Spanish class,
i.e. P(M) = 0.80 , P(S) = 0.60 and P(M∩S)=0.45
Required probability = P(neither M nor S)
= P(M'∩S')
= P(M∪S)' [∵P(A'∩B')=P(A∪B)']
=1- P(M∪S) [∵P(A')=1-P(A)]
= 1- (P(M)+P(S)- P(M∩S)) [∵P(A∪B)=P(A)+P(B)-P(A∩B)]
= 1- (0.80+0.60-0.45)
= 1- 0.95
= 0.05
hence, the probability that Helen does not get an A in either class= 0.05
Answer:
I don't see the following ratios.
Step-by-step explanation:
I think you should add the following.
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
#SPJ1
If i'm correct its c 1.5.
It should go base deci centi then milli
