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Aneli [31]
3 years ago
10

Randomly meeting a four child family with either one or exactly 2 boy children

Mathematics
1 answer:
Lelechka [254]3 years ago
3 0

Answer:

The probability of randomly meeting a four child family with either exactly one or exactly two boy children = (5/8) = 0.625

Step-by-step explanation:

Complete Question

The probability of randomly meeting a four child family with either exactly one or exactly two boy children.

Solution

The possible sample spaces for a family with four children include

4 boys and 0 Girl

BBBB

3 boys and 1 girl

BBBG BBGB BGBB GBBB

2 boys and 2 girls

BBGG BGBG BGGB GBBG GBGB GGBB

1 boy and 3 girls

BGGG GBGG GGBG GGGB

0 boy and 4 girls

GGGG

Total number of elements in the sample space = 16

Probability of an event is defined as the number of elements in that event divided by the Total number of elements in the sample space.

The required probability is a sum of probabilities.

The probability of meeting a four child family with exactly 1 boy = (4/16) = (1/4) = 0.25

The probability of meeting a four child family with exactly 2 boys = (6/16) = (3/8) = 0.375

The probability of randomly meeting a four child family with either exactly one or exactly two boy children = (4/16) + (6/16) = (10/16) = (5/8) = 0.25 + 0.375 = 0.625

Hope this Helps!!!

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The 5 th term of the given expression is 330(6x)^{7}( 6y)^{4}.

Step-by-step explanation:

Given,

(6x+6y)^{11}

To find the 5th term of the given expression.

Formula

In a binomial series (a+b)^{n}, the r th term is = nC(r-1) a^{n-r-1} b^{r-1}, where,

nC(r-1) = \frac{n!}{(r-1)!(n-r+1)!}

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Putting, n=11, r = r, a=6x and b=6y we get,

5 th term = 11C4(6x)^{7}( 6y)^{4}

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