Problem 1 Find the probability that in tossing a fair coin three times, there wi

Question

Problem 1 Find the probability that in tossing a fair coin three times, there will appear

a) 3 heads

b) 2 tails and 1 head

c) at least 1 head

d) not more than 1 tail

Problem 2

Find the probability that in a family of 4 children there will be

a) at least 1 boy

b) at least 1 boy and at least 1 girl

c) Out of 2000 families with 4 children each how many would you expect to have 1 or 2 girls. Assume that the probability of a male birth is 1/2.

Problem 3

Box I contains 3 red and 2 blue marbles while Box II contains 2 red and 8 blue marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box I; if it turns up tails, a marble is chosen from Box II.

a) Find the probability that a red marble is chosen.

b) Suppose that the person tossing the coin does not reveal whether it has turned up heads or tails (i.e., the box from which a marble was chosen is not revealed) but does reveal that a red marble was chosen. What is the probability that Box I was chosen (i.e., the coin turned up heads)?

Explanation / Answer

1) a) P(3 heads) = 0.53 = 0.125

b) P(2 tails and 1 head) = P(HTT) + P(THT) + P(TTH)

= 3 x 0.53 = 0.375

c) P(at least 1 head) = 1 - P(3 tails)

= 1 - 0.53

= 0.875

d) P(not more than 1 tail) = P(0 tail) + P(1 tail)

= 0.125 + 0.375

= 0.5

2) P(at least 1 boy) = 1 - P(4 girls)

= 1 - 0.54

= 1 - 0.0625

= 0.9375

b) P(at least 1 boy and 1 girl) = 1 - P(4 boys) - P(4 girls)

= 1 - 2x0.0625

= 0.8750

c) P(1 or 2 girls) = P(1 girl) + P(2 girls)

= 4C1 x 0.54 + 4C2x0.54

= 0.625

So, out of 2000 families, number of families with 1 or 2 girls = 0.625 x 2000 = 1250

NB: Please post different questions separately

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