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!!!
The probability of randomly meeting a four child family with either one or exactly 2 boy children is 0.625
The given parameters are:
[tex]\mathbf{n = 4}[/tex] --- number of children in a family
[tex]\mathbf{B = G = 0.5}[/tex] --- the probability of having a boy or a girl child
The distribution follows a binomial distribution.
Each probability is calculated using:
[tex]\mathbf{P(x) =^nC_x B^x G^{n-x}}[/tex]
For 1 or 2 boys, we have:
[tex]\mathbf{P(1\ or\ 2) = ^4C_1 \times 0.5^1 \times 0.5^{4-1} + ^4C_2 \times 0.5^2 \times 0.5^{4-2}}[/tex]
[tex]\mathbf{P(1\ or\ 2) = 4 \times 0.5^1 \times 0.5^3 + 6 \times 0.5^2 \times 0.5^2}[/tex]
[tex]\mathbf{P(1\ or\ 2) = 0.25 + 0.375}[/tex]
[tex]\mathbf{P(1\ or\ 2) = 0.625}[/tex]
Hence, the required probability is 0.625
Read more about probabilities at:
https://brainly.com/question/11234923
