On his drive to work, Leo listens to one of three radio stations A, B

On a station, the probability of getting song of choice is - 3/10.
The probability of not getting the song of choice on any station therefore is - 7/10.

So,
Probability of getting a song of choice on at least one of the stations

= 1-Probability of not getting a song of choice while trying all 3 stations.
= 1 - (7/10)*(7/10)*(7/10)
= 1- 343/1000
= 657/1000
=.657

Answer:- D

The probability that Leo will hear a song he likes on the way to work is the probability he will not turn off his radio. That is, either station A will be on for the entire trip, or station B or C will be on by the end of the trip.

The probability that station A will be on for the entire trip is 0.3.

Station B will be on by the end of the trip if station A did not play a song he likes AND station B did play a song he likes. The probability is 0.7 x 0.3 = 0.21.

Station C will be on by the end of the trip if station A did not play a song he likes AND station B did not play a song he likes AND station C did play a song he likes. The probability is 0.7 x 0.7 x 0.3 = 0.147.

Since these events are mutually exclusive, we add their probabilities, so the probability that a station will be on by the end of the trip is 0.3 + 0.21 + 0.147 = 0.657.

Answer: D

Can you explain why you are multiplying with 0.7 in second case and 0.7*0.7 in third case.

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Three mutually exclusive events can happen to Leo.

Event1 (E1): Leo hears song on A that he likes.

Event2 (E2): Leo does not like the song he hears on A and then turns to song on B that he likes.

Event3 (E3): Leo does not like the song he hears on A, so he turns to B and does not like the song he hears on B, and then finally turns to song on C that he likes.

We have to determine the probability (P) that Leo will hear a song that he likes. Note that since these three events are mutually exclusive, only one event can occur. P will therefore be the sum of the probabilities of each of the three events occurring.

We have:

P(Leo likes song on either of A, B, or C) = 0.3
P(Leo does not like song on either of A, B, or C) = 1-0.3 = 0.7

P(E1) = P(likes song on A) = 0.3

P(E2) = P(does not like song on A and then likes song on B) = 0.7*0.3

P(E3) = P(does not like song on A and then does not like song on B and then likes song on C) = 0.7*0.7*0.3

P = P(E1) + P(E2) + P(E3) = 0.657

ANSWER: D

NOTE: Recall the concepts for union of sets and intersection of sets . In the first, we add the probabilities. In the second, we multiply the probabilities.

We're considering 3 events that are exclusive; thus, we are adding the probability of the 3 events.

Probability of A = 0.3
Probability of not liking A but liking B = 0.7 * 0.3 = 0.21
Probability of not liking A OR B but liking C = 0.7 * 0.7 * 0.3 = 0.147

Probability of hearing a song Leo likes = {Probability of liking A} + {Probability of not liking A but liking B} + {Probability of not liking A or B but liking C} = 0.3 + 0.21 + 0.147 = 0.657

We can also consider the probability of something not happening and subtracting that value from 1.

1 - {Probability Leo doesn't like A} * {Probability Leo doesn't like B} * {Probability Leo doesn't like C} = 1 - 0.7 * 0.7 * 0.7 = 0.657

we jst use complement formula
P(E) = 1 - P(FAILURE EVENT)

The probability that neither of the stations is playing the song he likes: P=1−0.7∗0.7∗0.7=0.657P= 1−0.7∗0.7∗0.7=0.657

On his drive to work, Leo listens to one of three radio stations A, B or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns it to B. If B is playing a song he likes, he listens to it; if not, he turns it to C. If C is playing a song he likes, he listens to it; if not, he turns off the radio. For each station, the probability is 0.30 that at any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Leo will hear a song he likes?

Probability = Listen A + (Not Listen A)(Listen B) + (Not Listen A)(Not Listen B)(Listen C) = 0.3 + 0.7*0.3 + 0.7*0.7*0.3 = 0.300 + 0.210 + 0.147 = 0.657
or
Probability = 1 - (Not Listen A)(Not Listen B)(Not Listen C) = 1 - [0.7*0.7*0.7] = 0.657

Hence D

The probability that neither of the stations is playing the song he likes:

1 - [0.7*0.7*0.7] = 0.657

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