Questions: The probability that a person in the United States has type B'+ blood is 896 . Five unrelated people in the United States are selected at random. Complete parts (a) through (d) (a) Find the probability that all five have type B^+ blood. The probability that all five have type B^+ blood is 0.000003 (Round to six decimal places as needed.) (b) Find the probability that none of the five have type B'+ blood. The probability that none of the five have type B^+ blood is 0.659 (Round to three decimal places as needed.) (c) Find the probability that at least one of the five has type B^+ blood. The probability that at least one of the five has type B^+ blood is (Round to three decimal places as needed.)

The probability that a person in the United States has type B'+ blood is 896 . Five unrelated people in the United States are selected at random. Complete parts (a) through (d)
(a) Find the probability that all five have type B^+ blood.

The probability that all five have type B^+ blood is 0.000003
(Round to six decimal places as needed.)
(b) Find the probability that none of the five have type B'+ blood.

The probability that none of the five have type B^+ blood is 0.659
(Round to three decimal places as needed.)
(c) Find the probability that at least one of the five has type B^+ blood.

The probability that at least one of the five has type B^+ blood is (Round to three decimal places as needed.)
Transcript text: The probability that a person in the United States has type B' $^{+}$blood is 896 . Five unrelated people in the United States are selected at random. Complete parts (a) through (d) (a) Find the probability that all five have type $\mathbf{B}^{+}$blood. The probability that all five have type $\mathrm{B}^{+}$blood is 0.000003 (Round to six decimal places as needed.) (b) Find the probability that none of the five have type $B^{\prime}$ blood. The probability that none of the five have type $\mathrm{B}^{+}$blood is 0.659 (Round to three decimal places as needed.) (c) Find the probability that at least one of the five has type $\mathrm{B}^{+}$blood. The probability that at least one of the five has type $\mathrm{B}^{+}$blood is (Round to three decimal places as needed.)
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Solution

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Solution Steps

Step 1: Probability that All Five Have Type \( B^+ \) Blood

To find the probability that all five selected individuals have type \( B^+ \) blood, we use the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

  • \( n = 5 \) (the number of trials),
  • \( x = 5 \) (the number of successes),
  • \( p = 0.0896 \) (the probability of success),
  • \( q = 1 - p = 0.9104 \) (the probability of failure).

Calculating this gives:

\[ P(X = 5) = \binom{5}{5} \cdot (0.0896)^5 \cdot (0.9104)^0 = 1 \cdot 6e-06 \cdot 1 = 6e-06 \]

Thus, the probability that all five have type \( B^+ \) blood is \( 6e-06 \).

Step 2: Probability that None of the Five Have Type \( B^+ \) Blood

Next, we calculate the probability that none of the five individuals have type \( B^+ \) blood. Again, we use the binomial probability formula:

\[ P(X = 0) = \binom{5}{0} \cdot p^0 \cdot q^{5} = 1 \cdot 1 \cdot (0.9104)^5 \]

Calculating this gives:

\[ P(X = 0) = 1 \cdot 1 \cdot 0.625 = 0.625 \]

Thus, the probability that none of the five have type \( B^+ \) blood is \( 0.625 \).

Step 3: Probability that at Least One of the Five Has Type \( B^+ \) Blood

To find the probability that at least one of the five individuals has type \( B^+ \) blood, we can use the complement rule:

\[ P(\text{at least one}) = 1 - P(\text{none}) \]

Substituting the value we found in Step 2:

\[ P(\text{at least one}) = 1 - 0.625 = 0.375 \]

Thus, the probability that at least one of the five has type \( B^+ \) blood is \( 0.375 \).

Final Answer

  • Probability that all five have type \( B^+ \) blood: \( 6e-06 \)
  • Probability that none of the five have type \( B^+ \) blood: \( 0.625 \)
  • Probability that at least one of the five has type \( B^+ \) blood: \( 0.375 \)

\[ \boxed{ \begin{align_} P(\text{all five}) & = 6e-06 \\ P(\text{none}) & = 0.625 \\ P(\text{at least one}) & = 0.375 \end{align_} } \]

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