Binomial distribution?

Question

1) Joesph and 4 friends each have an independent probability 0.45 of winning a prize. Find the probability that

a)exactly 2 of the 5 friends win a prize.

b) Joseph and only one friend win a prize.

2) A bag contains 2 biased coins: coin A shows heads with probability 0.6, and coin B shows heads with probability 0.25. A coin is chosen at random from the bag, and tossed 3 times.

a) Find probability that 3 tosses of the coin show 2 heads and 1 tail in any order.

b)Find the probability that the coin chosen was coin A given that the 3 tosses result in 2 heads and 1 tail

Answer 1

1a) 5C2 * (0.45)^2 * (0.55)^3

b) (0.45) * 4C1 * (0.45)(0.55)^3

2a) P(2H1T) = (1/2) * 3 * (0.6)^2 (0.4) + (1/2) * 3 * (0.25)^2 * (0.75)

b) P(A|2H1T) = P(A and 2H1T)/P(2H1T), but

P(A and 2H1T) = 3 * (0.6)^2 (0.4) and P(2H1T) is found in 2a.

Answer 2

1.

a)

Pr (exactly 2 of the 5 friends win a prize)

= nCr (5,2) * (0.45)^2 * (0.55)^3

=0.336909375

b)

Pr (Joseph ∩ only one friend wins a prize)

= Pr (Joseph) * Pr (one friend wins a prize)

= [ 0.45 ] * [ nCr (4,1) * (0.45)^1 * (0.55)^3 ]

=0.13476375

2.

a)

Pr (3 tosses of the coin show 2 heads and 1 tail in any order)

= Pr (3 tosses show 2 heads and 1 tail in any order ∩ coin A) +Pr (3 tosses show 2 heads and 1 tail in any order ∩ coin B)

= [nCr (3,2) * (0.6)^2 * (0.4)^1] * 0.5 + [nCr (3,2) * (0.25)^2 * (0.75)^1] * 0.5

= 0.2863125

b)

Pr (coin chosen was coin A | 3 tosses result in 2 heads and 1 tail)

= (coin A ∩ 3 tosses result in 2 heads and 1 tail) / Pr(3 tosses result in 2 heads and 1 tail)

={[nCr (3,2) * (0.6)^2 * (0.4)^1] * [0.5]} / {[nCr (3,2) * (0.6)^2 * (0.4)^1] * 0.5 + [nCr (3,2) * (0.25)^2 * (0.75)^1] * 0.5}

= ~ 0.7544 (to 4 dp)