91影视

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

If the parachute opens for less than \({\rm{100}}\) , the equipment will be damaged. As a result, the chance of equipment damage can alternatively be expressed as \({\rm{P(X < 100)}}\).

Standardization provides:

\({\rm{X < 100}}\)

only if and only if

\(\begin{array}{*{20}{c}}{\frac{{{\rm{X - 200}}}}{{{\rm{30}}}}{\rm{ < }}\frac{{{\rm{100 - 200}}}}{{{\rm{30}}}}}\\{\frac{{{\rm{X - 200}}}}{{{\rm{30}}}}{\rm{ < }}\frac{{{\rm{ - 100}}}}{{{\rm{30}}}}}\\{{\rm{Z < - 3}}{\rm{.33}}}\end{array}\)

Thus

\({\rm{P(X < 100) = P(Z < - 3}}{\rm{.33)}}\)

\({\rm{Z}}\)represents a standard normal distribution \({\rm{rv}}\) with the \({\rm{cdf}}\) \({\rm{f(z)}}\) . Hence

\({\rm{P(X < 100) = P(Z < - 3}}{\rm{.33) = f( - 3}}{\rm{.33)}}\)

We start with Appendix Table A.3 to get \({\rm{f( - 3}}{\rm{.33)}}\).

\({\rm{f( - 3}}{\rm{.33) = 0}}{\rm{.0004}}\)

Hence

\({\rm{P(X < 100) = 0}}{\rm{.0004}}\)

Because this probability is equivalent to the chance that the parachute will sustain equipment damage,

\({\rm{P(\;parachute suffers damage\;) = 0}}{\rm{.0004}}\)

We can also say, using the complement rule of probability, that

\({\rm{P(\;parachute doesn't suffer damage\;) = 1 - 0}}{\rm{.0004 = 0}}{\rm{.9996}}\)

If an event A has a complement, then \({\rm{\bar A}}\) is the complement of that event.

\({\rm{P(\bar A) = 1 - P(A)}}\)

Assume that the payload of at least one of five independently dropped parachutes suffers equipment damage.

If we call the complement of event \({\rm{E by \bar E, then \bar E}}\) is the event in which none of the five independently dropped parachutes experience equipment damage.

Because all of the parachutes are released at the same time,

\(\begin{array}{*{20}{c}}{}&{{\rm{P(\bar E) = (P(parachute doesn't suffer damage)}}{{\rm{)}}^{\rm{5}}}}\\{}&{{\rm{P(\bar E) = (0}}{\rm{.9996}}{{\rm{)}}^{\rm{5}}}}\\{}&{{\rm{P(\bar E) = 0}}{\rm{.998}}}\end{array}\)

Thus,

\({\rm{P(E) = 1 - P(\bar E)P(E) = 1 - 0}}{\rm{.998P(E) = 0}}{\rm{.002}}\)

If events \({{\rm{A}}_{\rm{1}}}{\rm{,}}{{\rm{A}}_{\rm{2}}}{\rm{,}}{{\rm{A}}_{\rm{3}}}{\rm{ \ldots \ldots \ldots ,}}{{\rm{A}}_{\rm{n}}}\) are mutually independent, then the multiplication rule of probability applies.

\(P\left( {{A_1}{C}{A_2} {C}{A_3} {C} \ldots \ldots \ldots {C}{A_n}} \right) = P\left( {{A_1}} \right) \times P\left( {{A_2}} \right) \times P\left( {{A_3}} \right) \ldots \ldots ..P\left( {{A_n}} \right).\)