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Question
question 18 of 25
lan often takes his dog to the park. he estimates that 30% of the other dogs he sees are retrievers, 20% are terriers, and 20% are german shepherds. he designs a simulation.
let 0, 1, and 2 represent retrievers.
let 3 and 4 represent terriers.
let 5 and 6 represent german shepherds.
let 7, 8, and 9 represent other dogs.
the table shows the simulation results.
simulation results
01611 26343 87208 58889 58822
49003 49116 98444 34845 67970
63777 71890 01595 90500 43938
91971 58647 39440 28893 51995
according to this simulation, what is the probability that at least one of the next five dogs he sees is a german shepherd?
a. 0.65
b. 0.70
c. 0.75
d. 0.60
Step1: Count the total number of simulations
There are \(20\) simulations in total.
Step2: Count the number of simulations with no German shepherds
German shepherds are represented by \(5\) and \(6\). A simulation has no German shepherds if it has no \(5\)s or \(6\)s.
The simulations with no \(5\)s or \(6\)s are: \(01611\) (has a \(6\), so no), \(26343\) (has a \(6\), so no), \(87208\) (no \(5\)s or \(6\)s), \(58889\) (has a \(5\), so no), \(58822\) (has a \(5\), so no), \(49003\) (no \(5\)s or \(6\)s), \(49116\) (has a \(6\), so no), \(98444\) (no \(5\)s or \(6\)s), \(34845\) (has a \(5\), so no), \(67970\) (has a \(6\), so no), \(63777\) (has a \(6\), so no), \(71890\) (no \(5\)s or \(6\)s), \(01595\) (has a \(5\), so no), \(90500\) (has a \(5\), so no), \(43938\) (no \(5\)s or \(6\)s), \(91971\) (no \(5\)s or \(6\)s), \(58647\) (has a \(5\) and \(6\), so no), \(39440\) (no \(5\)s or \(6\)s), \(28893\) (no \(5\)s or \(6\)s), \(51995\) (has a \(5\), so no).
The number of simulations with no German shepherds is \(6\) (\(87208\), \(49003\), \(98444\), \(71890\), \(43938\), \(91971\), \(39440\), \(28893\)).
Step3: Calculate the probability of no German shepherds
The probability of no German shepherds \(P(\text{no GS})=\frac{6}{20}=0.3\)
Step4: Calculate the probability of at least one German shepherd
Using the formula \(P(\text{at least one GS}) = 1 - P(\text{no GS})\)
\(P(\text{at least one GS})=1 - 0.3=0.7\)
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B. \(0.70\)