A meeting consults was attended by 4 American and 5 Germans. if three consults were selected at random, construct the probability distribution of the random variable G representing the number of Germans

Answer:

Total of combinations of 3 consuls from among 4A+2G) = (4+2)!/(3!)(3!) = 20.

Possible combinations are: (0G+3A), (1G+2A), (2G+1A).

0G can occur in 2!/(2!)(0!) =1 way. 3A can occur in 4!/(1!)(3!) = 4 ways.

1G can occur in 2!/(1!)(1!) = 2 ways. 2A can occur in 4!/(2!)(2!) = 6 ways.

2G can occur in 2!/(0!)(2!) = 1way. 1A can occur in 4!/(3!)(1!) = 4 ways.

Thus, 3 consul distinct combinations are as follows:

0G’s appear in 1*4 = 4 of them.

1G appears in 2*6 = 12 of them.

2G’s appear in 1*4 = 4 of them.

A’s appear in all 20 possible combinations and G’s appear in just 16 of them.

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Pa brainliest

A meeting consults was attended by 4 American and 5 Germans. if three consults were selected at random, construct the probability distribution of the random variable G representing the number of Germans ​