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Luden [163]
3 years ago
5

A bag contains three red marbles, two green ones, one lavender one, four yellows, and five orange marbles. HINT [See Example 7.

How many sets of four marbles include all the red ones?
Mathematics
1 answer:
kirill115 [55]3 years ago
5 0

Answer:

There are 12 sets of four marbles include all the red ones.    

Step-by-step explanation:

Given : A bag contains three red marbles, two green ones, one lavender one, four yellows, and five orange marbles.

To find : How many sets of four marbles include all the red ones?

Solution :

Number of red marbles = 3

Number of green marbles = 2

Number of lavender marbles = 1

Number of yellow marbles = 4

Number of orange marbles = 5

We have to form sets of four marbles include all the red ones,

For position of getting red ones we have three red marbles i.e. ^3C_3

For the fourth one we have 12 choices i.e. ^{12}C_1

Total sets of four marbles include all the red ones is

=^3C_3\times ^{12}C_1

=1\times 12

=12

Therefore, There are 12 sets of four marbles include all the red ones.

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9966 [12]

a) i) Since blood types across individuals are independent, we have

Pr[1 = A and 2 = B] = Pr[1 = A] • Pr[2 = B]

where e.g. Pr[1 = A] means "probability that partner 1 has type A blood". According to the given distribution,

Pr[1 = A and 2 = B] = 0.37 • 0.13 = 0.0481 = 4.81%

a) ii) The event that partner 1 has type A and partner 2 has type B is exclusive from the event that partner 1 has type B and partner 2 has type A, and vice versa. In other words, only one of these events can happen, and not both simultaneously. This means the probability of either event occurring is equal to the sum of the probabilities of these events occurring individually. Then

Pr[(1 = A and 2 = B) or (1 = B and 2 = A)]

= Pr[1 = A and 2 = B] + Pr[1 = B and 2 = A]

Both probabilities are the same, and we found it in the previous part. So

Pr[(1 = A and 2 = B) or (1 = B and 2 = A)]

= 0.0481 + 0.0481 = 0.0962 = 9.62%

a) iii) The event that at least one partner has type O occurs if exactly one partner has type O, or both do. That is, "at least one partner with type O" contains 3 possible events,

• 1 = O and 2 = not O

• 1 = not O and 2 = O

• 1 = O and 2 = O

For the first event, by independence, is

Pr[1 = O and 2 = not O] = Pr[1 = O] • Pr[2 = not O]

and by definition of complementary event,

Pr[1 = O and 2 = not O] = Pr[1 = O] • (1 - Pr[2 = O])

From the given distribution, it follows that

Pr[1 = O and 2 = not O] = 0.44 • (1 - 0.44) = 0.2464 = 24.64%

The second event is the same as the first in terms of probability.

For the third event, we have

Pr[1 = O and 2 = O] = Pr[1 = O] • Pr[2 = O] = 0.44² = 0.1936 = 19.36%

Then the total probability of at least one O-type partner is

24.64% + 24.64% + 19.36% = 68.64%

b) Suppose partner 1 is type-B. Then partner 2 is an acceptable donor if they are either type-B or type-O. So we want to find

Pr[1 = O and (2 = B or 2 = O)]

By independence,

= Pr[1 = O] • Pr[2 = B or 2 = O]

By mutual exclusivity,

= Pr[1 = O] • (Pr[2 = B] + Pr[2 = O])

Then from the given distribution,

= 0.44 • (0.13 + 0.44) = 0.2508 = 25.08%

Now suppose partner 2 is type-B. Then partner 1 must be either type-B or type-O. But the math works out the same way, so that the overall probability is

25.08% + 25.08% = 50.16%

7 0
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