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Murrr4er [49]
3 years ago
5

A thief steals an ATM card and must randomly guess the correct five​-digit pin code from a 7​-key keypad. Repetition of digits i

s allowed. What is the probability of a correct guess on the first​ try?
Mathematics
1 answer:
bearhunter [10]3 years ago
3 0

Answer:

1 in 16,807 chance or a 0.00595% chance

Step-by-step explanation:

There is only one correct 5-digit pin that the thief must guess. If there are 7 possible values for each of the five digits, the total number of combinations is:

n = 7*7*7*7*7\\n= 16,807\ possibilities

Therefore, there is a 1 in 16,807 chance or a 0.00595% chance that the thief will guess on the first try.

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Please reference attached image for the problem that requires solving. Thank you so much for taking the time to help.
Lemur [1.5K]

Explanation:

The number of times the 6-sided number cube will be rolled will is

750

Let the numbers greater than 4 be represented below as

E_1\begin{gathered} E_1=\lbrace5,6\rbrace \\ n(E_1)=2 \end{gathered}

The number of sample space will be

n(S)=6

The probability of rolling a number greater than 4 will be calculated below as

\begin{gathered} Pr(E_1)=\frac{n(E_1)}{n(S)} \\ Pr(E_1)=\frac{2}{6}=\frac{1}{3} \end{gathered}

Hence,

To calculate the number of times a number greater than 4 will be rolled will be calculated below as

\begin{gathered} =Pr(E_1)\times750 \\ =\frac{1}{3}\times750 \\ =250times \end{gathered}

Hence,

The final answer is

\Rightarrow250\text{ }times

5 0
1 year ago
Two of the 240 passengers are chosen at random. Find the probability that
hjlf

Step-by-step explanation:

there are in total 240 passengers.

out of these 240, there are 150+30=180 passengers that are in holiday.

and 240-180 = 60 passengers are not.

if we pick one passenger then the probability is 180/240 = 3/4 = 0.75 that he/she is on holiday.

remember : desired "events" over total "events".

i)

now we pick 2 passengers.

the probabilty for the first one to be on holiday is again

3/4 or 0.75.

if that event happens, then we have only 179 passengers out of now 239 to be on holiday.

and to pick one out of that pool to be on holiday is then

179/239 = 0.748953975...

and for both events to happen in one scenario we need to multiply both probabilities (it is an "and" relation, while an addition would be for an "exclusive or" relation).

the probabilty that we pick 2 passengers on holiday is

3/4 × 179/239 = 0.561715481... ≈ 0.5617

we cannot simply square the basic probability of 0.75 (0.75² = 0.5625), because that would mean we pick one passenger, then put him back into the crowd, and then pick a second time (with a chance to pick the same person again). like with rolling a die.

but that is not the scenario as I understand it. it is to pick a passenger, then keep that person singled out and pick a second passenger. hence the difference.

ii)

exactly one of the two is in holiday.

that means

either the first one is on holiday and the second one is not, or the the second one is and the first one is not.

now we model this logic statement in probabilty arithmetic.

please note that after the first pull we need to update the numbers for the remaining pool depending on the result of the first pull.

the total remaining is in both cases 239. but either the remaining people on holiday go down to 179 (and not in holiday stays 60), or the remaining people not on holiday go down to 59 (and on holiday stays 180).

so, the first one is on holiday, and the second one is not :

3/4 × 60/239 (remember : "and" relation)

= 3 × 15/239 = 45/239 = 0.188284519...

the first one is not on holiday, and the second one is :

60/240 × 180/239 = 1/4 × 180/239 = 45/239 =

= 0.188284519...

since there is no overlap of the potential events (there is no event that could be in both cases), this is an exclusive or relation, and we can add the probabilities.

so, the probability for exactly one of the picked passengers to be on holiday is

2×0.188284519... = 0.376569038... ≈ 0.3766

8 0
2 years ago
3. In the fig. AD = DC and AB = BC. Prove that ΔADB = ΔCDB<br>​
xxMikexx [17]

Answer:

The two column proof is presented as follows;

The given parameters are;

\overline {AD} = \overline {DC} and \overline {AB} = \overline {BC}

Statement               {}                     Reason

\overline {AD} = \overline {DC} and \overline {AB} = \overline {BC}       {}       Given

\overline {BD}  ≅  \overline {BD}               {}                    Reflexive property

\overline {BD}  =  \overline {BD}               {}                     By the definition of congruency

ΔADB ≅ ΔCDB             {}                By Side-Side-Side (SSS) rule of congruency

ΔADB = ΔCDB              {}                 By the definition of congruency

If the three sides of one triangle are congruent to the corresponding three sides of another triangle, then both triangles are said to be congruent according to the SSS rule of congruency

Step-by-step explanation:

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3 years ago
A leg of a right triangle measures 63 yards,
erma4kov [3.2K]

Answer:

16

Step-by-step explanation:

7 0
2 years ago
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5ab^2 - 12a^2b + 3ab + ab^2 + 4a^2b
Natali [406]
By combining like terms, the answer is a
8 0
2 years ago
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