the probability that a person in the united states has type $b^+$ blood is 10%. three unrelated people in the united states are selected at random. complete parts (a) through (d).

(a) find the probability that all three have type $b^+$ blood.
the probability that all three have type $b^+$ blood is 0.001.
(round to six decimal places as needed.)

(b) find the probability that none of the three have type $b^+$ blood.
the probability that none of the three have type $b^+$ blood is .
(round to three decimal places as needed.)

Question

the probability that a person in the united states has type $b^+$ blood is 10%. three unrelated people in the united states are selected at random. complete parts (a) through (d).

(a) find the probability that all three have type $b^+$ blood.
the probability that all three have type $b^+$ blood is 0.001.
(round to six decimal places as needed.)

(b) find the probability that none of the three have type $b^+$ blood.
the probability that none of the three have type $b^+$ blood is  .
(round to three decimal places as needed.)

Accepted Answer

{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "blank_filling_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Independent Events Probability", "Multiplication Rule for Probability", "Complementary Events" ], "new_concepts": [], "current_concepts": [ "Independent Events Probability", "Multiplication Rule for Probability", "Complementary Events" ] } ## Define given values and events Using the *Independent Events Probability* knowledge point $$ \begin{aligned} &P(B^+) = 0.10\ &n = 3 \text{ independent people} \end{aligned} $$ ## Calculate probability for part (a) Using the *Multiplication Rule for Probability* knowledge point $$ \begin{aligned} P(\text{all three } B^+) &= P(B^+) \times P(B^+) \times P(B^+)\ &= 0.10^3\ &= 0.001000 \end{aligned} $$ ## Find the complement probability Using the *Complementary Events* knowledge point $$ \begin{aligned} P(\text{not } B^+) &= 1 - P(B^+)\ &= 1 - 0.10\ &= 0.90 \end{aligned} $$ ## Calculate probability for part (b) Using the *Multiplication Rule for Probability* and *Independent Events Probability* knowledge points $$ \begin{aligned} P(\text{none of the three } B^+) &= P(\text{not } B^+)^3\ &= 0.90^3\ &= 0.729 \end{aligned} $$ ## Question 1 (a) Find the probability that all three have type $\text{B}^+$ blood. The probability that all three have type $\text{B}^+$ blood is $0.001000$. (Round to six decimal places as needed.) ## Question 2 (b) Find the probability that none of the three have type $\text{B}^+$ blood. The probability that none of the three have type $\text{B}^+$ blood is $0.729$. (Round to three decimal places as needed.) { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Statistics", "Independent Events Probability" ] }

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QUESTION IMAGE

Question

Explanation:

Define given values and events

Calculate probability for part (a)

Find the complement probability

Calculate probability for part (b)

Answer:

Question 1

Question 2