Question

(a) When you toss an unbiased coin five times, what is the probability that you will obtain exactly 3 heads, and 2 tails? 

(b) In (a), what is the probability that you will obtain exactly 3 heads, and 2 tails, in that order? 

(c) When you spin an unbiased die, there are six possible outcomes. What is the probability that you spin an unbiased die once, and you get both a 2, and a 6? What is the probability that in one spin, you get a 2, or a 6? What laws of probability holds true in this case?

Answer 1

a)here as 3 heads and 2 tails can be arranged =(5!/(3!*2!)) orders and each outcome has probability =1/2

therefore P(getting 3 heads and 2 tails) =(5!/(3!*2!))*(1/2)⁵ =5/16

b)

as there is only one possible order and each outcome has probability =1/2

therefore P(3 heads and 2 tails in this order)=(1/2)⁵=1/32

c)

P(getting a 2 and a 6 on one spin) =0 (as there can be only one outcome possible on one spin; therfore both 2 and 6 can not appear at the same time as they are mutually exclusive events)

P(getting a 2 or 6) =P(getting a 2)+P(getting a 6)=(1/6)+(1/6)=1/3

here addition rule of probability is applicable   (for 2 mutually exclusive events)

Answer 2

Heads only