Question

Suppose you just purchased a digital music player and have put 15 tracks on it. After listening to them you decide that you like 4 of the songs. With the random feature on your​ player, each of the 6 songs is played once in random order. Find the probability that among the first two songs played ​(a) You like both of them. Would this be​ unusual? ​(b) You like neither of them. ​(c) You like exactly one of them. ​(d) Redo​ (a)-(c) if a song can be replayed before all 6 songs are played.

Answer 1

Answer:

(a) 0.3084

(b)0.6916

(c)0.3394

d 0.2804, 0.7196, 0.3856

Step-by-step explanation:

p=4/15, q= 1 - 4/15=11/15

(a) n= 2 , x = 2.

We use the Bernoulli random distribution formula.

P(X= 2) = 6C2(4/15)²(11/15)⁴

=15 ×0.0711 × 0.2892

= 0.3084

(b) Pr(X=2) + P(X=2)' = 1

P(X=2)' = 1 -P(X=2)

= 1 - 0.3084

= 0.6916

(c) n= 6, X=1

Pr(X=1) = 6C1 (4/15)¹(11/15)^5

= 6 × (4/15) × (11/15)^5

= 6 × 0.2667 × 0.2121

= 0.3394

(d) This means only 5 is replayes in the random order.

d(a) n= 5, X= 2

Pr(X=2) = 5C2(4/15)²(11/15)³

= 10 × 0.0711× 0.3944

= 0.2804

d(b) Pr(X=2) + P(X=2)' = 1

P(X=2)' = 1 - 0.2804

P(X=2)' = 0.7196

d(c) Pr(X=1) = 5C1 (4/15)¹(11/15)⁴

P(X=1) = 5 × 0.2667 × 0.2892

= 0.3856

Answer 2

Answer:

(a) 0.4 (b) 0.067 (c) 0.53 (d) 0.44; 0.11; 0.45

Step-by-step explanation:

(a) In this part, you like 4 out of 6, the probability of choosing one at random will be 4/6 = 2/3. If you choose another one at random, the probability will be 3/5.  Thus, P(like both) = (2/3)*(3/5) = 0.4.  No, because it is above 0.05.

(b) The probability that you do not like a random selection is 2/6. The probability that you do not like the second random selection is 1/5. The probability of not liking both will be (2/6)*(1/5) = 0.067

(c) Based on the available options that are like both, do not like both, and like only one, we can deduce the probability of liking exactly one as:

1 - P(like both) - P(like neither) = 1 - 0.4 - 0.067 = 0.53  

(d) If the songs can be repeated:

P(like both) = (2/3)(2/3) = 0.44. It is not unusual because it is more than 5%.

P(like neither) = (2/6)(2/6) = 0.11

P(like exactly one) = 1 - 0.44 - 0.11 = 0.45