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Bunuel
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3 friends A, B and C started driving from point P towards point Q in 12 Oct 2021, 08:00
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66% (02:20) correct 34% (02:26) wrong based on 224 sessions

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3 friends A, B and C started driving from point P towards point Q in their individual cars. The individual probability for A to reach point Q first was 0.6. For B and C, the probabilities to reach point Q first were 0.7 and 0.4 respectively. What is the probability that either B or C, but not A, reached point Q first?


A. 0.048
B. 0.108
C. 0.168
D. 0.216
E. 0.324
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D
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3 friends A, B and C started driving from point P towards point Q in 12 Oct 2021, 08:11
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----P to reach---P to not reach
A--0.6---------0.4
B--0.7----------0.3
C--0.4----------0.6

The probability that either B or C, but not A, reached point Q first:
0.7*0.6*0.4+0.4*0.3*0.4
0.216 option D


Bunuel
3 friends A, B and C started driving from point P towards point Q in their individual cars. The individual probability for A to reach point Q first was 0.6. For B and C, the probabilities to reach point Q first were 0.7 and 0.4 respectively. What is the probability that either B or C, but not A, reached point Q first?


A. 0.048
B. 0.108
C. 0.168
D. 0.216
E. 0.324
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3 friends A, B and C started driving from point P towards point Q in 13 Jul 2024, 04:56
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Archit3110
----P to reach---P to not reach
A--0.6---------0.4
B--0.7----------0.3
C--0.4----------0.6

The probability that either B or C, but not A, reached point Q first:
0.7*0.6*0.4+0.4*0.3*0.4
0.216 option D


Bunuel
3 friends A, B and C started driving from point P towards point Q in their individual cars. The individual probability for A to reach point Q first was 0.6. For B and C, the probabilities to reach point Q first were 0.7 and 0.4 respectively. What is the probability that either B or C, but not A, reached point Q first?


A. 0.048
B. 0.108
C. 0.168
D. 0.216
E. 0.324
Show more
­IMO, It should be [P(B) +P(C) - P(B U C)] x [1-P(A)]. If its either B or C that can reach 1st then, shouldn't we subtract the case where both P and Q reach together? 

Correct me if I'm wrong and please elaborate on the solution. Archit3110 Bunuel
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3 friends A, B and C started driving from point P towards point Q in 13 Jul 2024, 09:09
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Budhaditya_Saha

Archit3110
----P to reach---P to not reach
A--0.6---------0.4
B--0.7----------0.3
C--0.4----------0.6

The probability that either B or C, but not A, reached point Q first:
0.7*0.6*0.4+0.4*0.3*0.4
0.216 option D


Bunuel
3 friends A, B and C started driving from point P towards point Q in their individual cars. The individual probability for A to reach point Q first was 0.6. For B and C, the probabilities to reach point Q first were 0.7 and 0.4 respectively. What is the probability that either B or C, but not A, reached point Q first?


A. 0.048
B. 0.108
C. 0.168
D. 0.216
E. 0.324
­IMO, It should be [P(B) +P(C) - P(B U C)] x [1-P(A)]. If its either B or C that can reach 1st then, shouldn't we subtract the case where both P and Q reach together? 

Correct me if I'm wrong and please elaborate on the solution. Archit3110 Bunuel
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In this case, we are talking about arriving first so you cannot have the event of B and C reaching together first­
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3 friends A, B and C started driving from point P towards point Q in 13 Jul 2024, 18:12
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Okay. Then what wrong am I doing wrong if I equate it like this?

[P(B) + P(C)]x[1-P(A)].

This translates to probability of B reaching first OR(addition) C reaching first AND(multiplication) A not reaching first.
?? pierren Bunuel

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3 friends A, B and C started driving from point P towards point Q in 13 Jul 2024, 20:46
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P(B or C but not A)=P(B reaches first and not A)+P(C reaches first and not A)
P(B reaches first and not A)=0.7×(1−0.6)=0.7×0.4=0.28
P(C reaches first and not A)=0.4×(1−0.6)=0.4×0.4=0.16
P(B or C but not A)=0.28+0.16=0.44
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3 friends A, B and C started driving from point P towards point Q in 13 Jul 2024, 20:52
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Aarif2204
P(B or C but not A)=P(B reaches first and not A)+P(C reaches first and not A)
P(B reaches first and not A)=0.7×(1−0.6)=0.7×0.4=0.28
P(C reaches first and not A)=0.4×(1−0.6)=0.4×0.4=0.16
P(B or C but not A)=0.28+0.16=0.44
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Exactly this is the method that I have used. But the answer is probably wrong. But I’m not able to understand where have I mistaken?

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3 friends A, B and C started driving from point P towards point Q in 14 Jul 2024, 00:39
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­Let A = 0.6, B = 0.7, C = 0.4

Let ~A = 0.4, ~B = 0.3, ~C = 0.6

From the ques: 

we need to find (~A)[ B * ~C + ~B * C]

= (0.4)[0.7*0.6 + 0.3*0.4]

= 0.4 * 54/100

= 216/1000 = 0.216 
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3 friends A, B and C started driving from point P towards point Q in Updated on: 14 Jul 2024, 05:26
Originally posted by Shrvan7h on 14 Jul 2024, 02:44.
Last edited by Shrvan7h on 14 Jul 2024, 05:26, edited 1 time in total.
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Why did we discount the possibility of both B and C reaching first, and A not reaching first? [I meant this P(B∩C∩A')]

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3 friends A, B and C started driving from point P towards point Q in 14 Jul 2024, 05:11
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Budhaditya_Saha
Okay. Then what wrong am I doing wrong if I equate it like this?

[P(B) + P(C)]x[1-P(A)].

This translates to probability of B reaching first OR(addition) C reaching first AND(multiplication) A not reaching first.
?? pierren Bunuel

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­In fact, if I am right, you are trying to translate the event of (B U C) which again is not possible as two people can not arrive first at the same time.
However, if you want to translate this into a formula, it is P( B U C) = P(B) + P(C) - P(B inter C)
so you would have [P(B) + P(C) - P(B inter C)] * [1 - P(A)], what is wrong in our case with the given meaning of B U C.

I did the same error and corrected myself by thinking with the order, here we need and arrangment / order  as we are talking about an order of arrival.

So the formula is P(B) * [1 - P(C)] * [1 - P(A)] OR [1 - P(B)] * P(C) * [1 - P(A)]­ which means B arrives first and not C and not A OR C arrives first and not B and not A
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3 friends A, B and C started driving from point P towards point Q in 21 Sep 2025, 05:13
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