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grade | p(grade) a | 0.1 b | 0.3 c | 0.35 d | 0.05 f |? 2a. if i random…

Question

grade | p(grade)
a | 0.1
b | 0.3
c | 0.35
d | 0.05
f |?
2a. if i randomly choose 1 student at random from the distribution, what is the probability he/she got an f? round to the tenth.
0.2
2b. if i randomly choose 1 student at random from the distribution, what is the probability he/she did not get an a or b? round to the tenth.
0.6
2c. if i randomly choose 1 student at random from the distribution, what is the probability he/she got an a and a b?

Explanation:

Step1: Calculate probability of F

The sum of all probabilities in a distribution is \(1\). Let \(P(F)\) be the probability of getting an \(F\).
We know that \(P(A)+P(B)+P(C)+P(D)+P(F) = 1\).
Substituting the given values: \(0.1 + 0.3+0.35 + 0.05+P(F)=1\).
Simplifying the left - hand side: \(0.8 + P(F)=1\).
Solving for \(P(F)\): \(P(F)=1 - 0.8=0.2\).

Step2: Calculate probability of not A or B

The probability of getting an \(A\) or \(B\) is \(P(A\cup B)=P(A)+P(B)\) (since \(A\) and \(B\) are mutually exclusive events).
\(P(A)+P(B)=0.1 + 0.3 = 0.4\).
The probability of not getting an \(A\) or \(B\) is \(P(\overline{A\cup B})=1 - P(A\cup B)\).
So \(P(\overline{A\cup B})=1 - 0.4 = 0.6\).

Step3: Calculate probability of A and B

Since a student cannot get both an \(A\) and a \(B\) simultaneously (mutually exclusive events), \(P(A\cap B) = 0\).

Answer:

2a. \(0.2\)
2b. \(0.6\)
2c. \(0\)