events. The probability of first one head and then nine tails is

equal to:

P(H,9T)=.7*.3^9 .

Similarly, we calculate probability that we are getting exactly

one head on the second toss:

P(T,H,8T)= (.3)(.7)(.3)^8 =.7*.3^9. We similarly get .7*.3^9 for

just one head in each of the other 8 possible positions. To

find the probability of the exactly one head (in all possible

arrangements with nine tails) we have to add up these ten

equal probabilities:

P(exactly one head)=10/(.7*.3^9).

b) Let us first calculate the probability of getting either no

heads or 1 heads. The probability of getting no heads is .

3^10. The probability of getting 1 heads is (as we showed

above) 10/(.7*.3^9). Thus the probability of getting either 0 or

1 heads is

.3^10+ 10/(.7*.3^9)

Since the probability of getting either 0 or 1 heads or at least

2 heads = 1, we get the equation .3^10+ 10/(.7*.3^9) + P(at

least 2 heads) =1, or transposing

P(at least 2 heads) = 1 - .3^10 - 10/(.7*.3^9)

11) a)The probability of such event is one specific choice

over the total number of possibilities. The total number would

be 14 for president, times 13 for vice president, times 12 for

treasurer, times 11 for the secretary:

P(Esther-P, Chana-VP, Rivka-T, Sarah-S)=1/(14*13*12*11)

=1/24024

b) The total number of arrangements of 4 people is 4!=12,

the probability is

P(Esther, Chana, Rivka, Sarah to be P,VP,T,S)

=12/24024=1/2002

12) The probability of 4 kings in the hand of 12 can be found

as a ratio of all hands with 4 kings (and 8 other cards) over

the total number of hands of 12.

The total number of hands of 12 would be found in the

following manner: the first card is picked up out of 36, times

the second card picked up out of 35, and so on: 36*35*…*25.

This would be the total number of hands if order matters, but

in our example order does not matter, so we have to divide

the previous result by 12! – the total number of possible

arrangements of 12 cards. In a hand with 4 kings we have

chosen 4 cards, the remaining 8 positions would be occupied

in the following ways: 32*31*30*29*28*27*26*25/8!-(it is

divided by 8! because order does not matter).

After canceling out the same numbers in the numerator and

denominator, our probability of getting 4 kings in a hand of 12

will be:

P(4K from 12)= 12*11*10*9/(36*35*34*33)=1/(7*17)=1/119.