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.

Question

Mathematics

Chloe Obrien

25 October, 21:39

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.

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Answers (2)

Know the Answer?

Alondra Beck · Answer 1

(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

Lana Ibarra · Answer 2

(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