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cluponka [151]
3 years ago
12

4) (20 pts) A box contains 5 apples and 6 oranges. Four children each receive a fruit from the box, one after the other, randoml

y chosen, without replacement. What is the probability that all four children receive the same fruit
Mathematics
1 answer:
andrezito [222]3 years ago
7 0

Answer:

Probability of an event = Number of outcomes favourable to that event/ Number of all possible outcomes.

The set of all possible outcomes is called the sample space and is denoted by S.

Any 3 fruits out of 11 in the box can be randomly selected in 11C3= (11x10 x 9)/(1 x 2x3) = 165 ways. Hence n(S) = 165.

We want the desired event E that the 1 fruit of each type is chosen. Now 1 apple out of 5 can be chosen in 5C1 = 5 ways, 1 out of 4 oranges in 4C1 = 4 and 1 banana out of 2 in 2C1 = 2 ways. Therefore, by fundamental counting principle, one of each type of fruit can be chosen in 5x4x2 = 40 ways So n(E) = 40.

Hence the probability of getting three fruits one of each type is = n(E)/ n(S) = 40/165 = 8/33.

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spin [16.1K]

Answer:

c = 7

d = 5

Step-by-step explanation:

Notice that in the first expression, x^c is inside a square root, and only perfect squares can be extracted from it. On the simplified form shown on the right hand side, we have x^3 outside the root and a single "x" left inside. In order for such to happen (x^3 get outside the root) there must have been an x^6 inside the square root. This together with the sole "x" that was left in the root, totals seven factors of x that should have been originally inside the square root:

x^6 * x = x^7  therefore c was a "7"

In the second expression we have a CUBIC root, so only perfect cubes can get extracted from it. Since there is one factor "x" shown in the simplified form (right hand side of the equal sign), that means that it must have been an x^3 (perfect cube) apart from the x^2 that was left inside the root. This makes the original power of x to be a 3 + 2 = 5.

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(8 2/5, 8 2/5)
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4 years ago
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