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According to a report from a particular university, 46% of female undergraduates take on debt. Find the probability that none of the female undergraduates have taken on debt if 9female undergraduates were selected at randomWhat probability should be found?MA PO female undergraduates take on debt)OB P(9 female undergraduates take on debt)OC P(1 female undergraduate takes on debt)OD P(2 female undergraduates take on debt)The probability that none of the female undergraduates take on debt is I(Type an integer or decimal rounded to three decimal places as needed)1}0vo(0.MorexHelp Me Solve ThisView an ExampleGet More HelpClear AllCheck Answer

Sagot :

For this exercise we use the probability function of the binomial distribution, also called the Bernoulli distribution function, is expressed with the formula:

[tex]P(x)=\frac{n!}{(n-x)!\cdot x!}\cdot p^x\cdot q^{n-x}[/tex]

Where:

• n, = the number of trials

,

• x, = the number of successes desired

,

• p,= probability of getting a success

,

• q, = probability of getting failure

From the exercise we can identify:

[tex]\begin{gathered} n=9 \\ x=0 \\ p=0.46 \\ q=1-p \\ q=0.54 \end{gathered}[/tex]

Replacing in the equation of the binomial distribution:

[tex]\begin{gathered} P(0)=\frac{9!}{(9-0)!\cdot0!}\cdot(0.46)^0\cdot(0.54)^{9-0} \\ P(0)=0.0039 \\ P(0)=0.004 \end{gathered}[/tex]

The answer is P(0 female undergraduates tak on debt)