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3 years ago

A two-digit locker combination has two nonzero digits and no digit is repeated in any combination.

Mathematics

1 answer:

zzz [600]3 years ago

7 0

Answer:

Answer choice C

Step-by-step explanation:

Similarly to the previous problem, there are two possibilities. In the first possibility, the chance of picking a digit lower than 7 is 6/9 or 2/3, because there are 6 digits below 7 and 9 digits in total. In this scenario, the probability of the second digit being below 7 is 5/8, since one of the digits has already been taken. The probability here is 5/8*6/9=5/12. In the other scenario, the probability of picking a number higher than 7 is 3/9 or 1/3. In this scenario, the probability of picking below 7 for the second digit is 6/8 or 3/4, because one of the digits has not been taken. 3/9*6/8= 1/4. Adding these two together, you get 8/12=2/3, or answer choice C. Hope this helps!

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Lelechka [254]

Answer:

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = 0.2737

d) μ = 15*0.75 = 11.25. σ² = 11.25*0.25 = 2.8125

Step-by-step explanation:

X is a binomial random variable with parameters n = 15, p = 0.75. Therefore

$P(X=k) = {15 \choose k} * 0.75^{k}*0.25^{15-k}$

For any integer k between 0 and 15, and 0 for other values of k.

P(X>10) = P(X=11) + P(X=12)+ P(X=13)+P(X=14)+P(x=15)

$P(X=11) = {15 \choose 11} * 0.75^{11} * 0.25^4 = 0.2252$

$P(X=12) = {15 \choose 12} * 0.75^{12} * 0.25^3 = 0.2252$

$P(X=13) = {15 \choose 13} * 0.75^{13} * 0.25^2 = 0.1559$

$P(X=14) = {15 \choose 14} * 0.75^{14} * 0.25 = 0.0668$

$P(X=15) = {15 \choose 15} * 0.75^{15} = 0.0134$

Thus,

$P(X>10) = 0.2252+ 0.2252+ 0.1559+0.0668+0.0134 = 0.6865$

c) P(6 ≤ X ≤ 10) = P(X = 6) + P(X = 7) + P(X = 8) + P(X=9) + P(X=10)

$P(X=6) = {15 \choose 6} * 0.75^{6} * 0.25^9 = 0.0034$

$P(X=7) = {15 \choose 7} * 0.75^{7} * 0.25^8 = 0.0131$

$P(X=8) = {15 \choose 8} * 0.75^{8} * 0.25^7 = 0.0393$

$P(X=9) = {15 \choose 9} * 0.75^{9} * 0.25^6 = 0.0918$

$P(X=10) = {15 \choose 10} * 0.75^{10} * 0.25^{5} = 0.1652$

Thereofre,

$P(6 \leq X \leq 10) = 0.0034 + 0.0134 + 0.0393 + 0.0918 + 0.1652 = 0.2737$

d) μ = n*p = 15*0.75 = 11.25

σ² = np(1-p) = 11.25*0.25 = 2.8125

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3 years ago

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Answer:

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Step-by-step explanation:

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Answer:

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Answer:

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Step-by-step explanation:

△

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Step-by-step explanation:

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