PP2 PS 63/89/92

[vantreal]

63. 7231-!-item-!-187;#058&005082
The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200, inclusive?

(A) 5,100
(B) 7,550
(C) 10,100
(D) 15,500
(E) 20,100

Ans: (B)


89. 9835-!-item-!-187;#058&007168
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

(A) (0.6)^5
(B) 2(0.6)^4
(C) 3(0.6)^4(0.4)
(D) 4(0.6)^4(0.4) + (0.6)^5
(E) 5(0.6)^4(0.4) + (0.6)^5

Ans: (E)


92. 10192-!-item-!-187;#058&007298
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 to B. If B is playing a song he likes, he listens to it; if not, he turns 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?

(A) 0.027
(B) 0.090
(C) 0.417
(D) 0.657
(E) 0.900

Ans: (D)

謝謝大家的幫忙~

[Enriqueta]

[(102+200)*50] / 2 = 7550

63. 7231-!-item-!-187;#058&005082
The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200, inclusive?

(A) 5,100
(B) 7,550
(C) 10,100
(D) 15,500
(E) 20,100

Ans: (B)

[Enriqueta]

89. 出現5次都是人頭的機率=(0.6)^5
出現4次人頭1次字的機率=5*(0.6)^4(0.4)，乘以5是因為總共丟5次，字可能在第一次或第二次或第三次…出現，有5種可能，所以乘以5。
2個加起來就是答案E。

92.
在A電台就有喜歡的歌的機率是0.3
在B電台的機率是0.7*0.3 (因為要先在A不喜歡，才會在B喜歡，所以是0.7*0.3)
在C電台的機率是0.7*0.7*0,3
3個機率加起來就是0.657

89. 9835-!-item-!-187;#058&007168
For one toss of a certain coin, the probability that the outcome is heads is 0.6. If this coin is tossed 5 times, which of the following is the probability that the outcome will be heads at least 4 times?

(A) (0.6)^5
(B) 2(0.6)^4
(C) 3(0.6)^4(0.4)
(D) 4(0.6)^4(0.4) + (0.6)^5
(E) 5(0.6)^4(0.4) + (0.6)^5

Ans: (E)


92. 10192-!-item-!-187;#058&007298
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 to B. If B is playing a song he likes, he listens to it; if not, he turns 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?

(A) 0.027
(B) 0.090
(C) 0.417
(D) 0.657
(E) 0.900

Ans: (D)

謝謝大家的幫忙~

[vantreal]

[(102+200)*50] / 2 = 7550

謝謝您的回覆,
不過我想問一下那題目前面一句 the sum of the first 50 consecutive even integers is 2550, 和解答有關係嗎?
因為好像不用這句也是可以解答, 是不是呢?

[Enriqueta]

我是覺得沒影響啦，它大概只是在讓我們確認這個公式的正確性吧。
[(2+100)*50]/2 =2550

[(102+200)*50] / 2 = 7550

謝謝您的回覆,
不過我想問一下那題目前面一句 the sum of the first 50 consecutive even integers is 2550, 和解答有關係嗎?
因為好像不用這句也是可以解答, 是不是呢?

[小花]

92. 10192-!-item-!-187;#058&007298
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 to B. If B is playing a song he likes, he listens to it; if not, he turns 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?

(A) 0.027
(B) 0.090
(C) 0.417
(D) 0.657
(E) 0.900

Ans: (D)

前輩的解釋:
在A電台就有喜歡的歌的機率是0.3
在B電台的機率是0.7*0.3 (因為要先在A不喜歡，才會在B喜歡，所以是0.7*0.3)
在C電台的機率是0.7*0.7*0,3
3個機率加起來就是0.657

從題目給的資訊我們怎麼知道要把三首歌的機率相加呢?

[emmaaqua]

92. 10192-!-item-!-187;#058&007298
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 to B. If B is playing a song he likes, he listens to it; if not, he turns 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?

(A) 0.027
(B) 0.090
(C) 0.417
(D) 0.657
(E) 0.900

Ans: (D)

前輩的解釋:
在A電台就有喜歡的歌的機率是0.3
在B電台的機率是0.7*0.3 (因為要先在A不喜歡，才會在B喜歡，所以是0.7*0.3)
在C電台的機率是0.7*0.7*0,3
3個機率加起來就是0.657

從題目給的資訊我們怎麼知道要把三首歌的機率相加呢?

我想是因為Leo不是聽A電台就是BorC電台
三者會發生的機率是各自獨立的
所以是"相加"

[little8]

89. 出現5次都是人頭的機率=(0.6)^5
出現4次人頭1次字的機率=5*(0.6)^4(0.4)，乘以5是因為總共丟5次，字可能在第一次或第二次或第三次…出現，有5種可能，所以乘以5。
2個加起來就是答案E。

請教出現4次人頭1次字的機率=5*(0.6)^4(0.4)
其中,為什麼是(0.6)^4(0.4)?
不是應該是(0.6)^4+5(0.4)?

煩請解惑哩~~

[Huang Hsin-Yi]

同樓上問

[william0625]

89. 出現5次都是人頭的機率=(0.6)^5
出現4次人頭1次字的機率=5*(0.6)^4(0.4)，乘以5是因為總共丟5次，字可能在第一次或第二次或第三次…出現，有5種可能，所以乘以5。
2個加起來就是答案E。

請教出現4次人頭1次字的機率=5*(0.6)^4(0.4)

其中,為什麼是(0.6)^4(0.4)?

不是應該是(0.6)^4+5(0.4)?



煩請解惑哩~~

5*(0.6)^4(0.4)為同一事件
(0.6)^4+5(0.4)為兩個獨立事件