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Three men and 2 women will present 5 consecutive speeches, 1 by each p

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Three men and 2 women will present 5 consecutive speeches, 1 by each p  [#permalink]

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New post 22 Jul 2016, 11:23
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Three men and 2 women will present 5 consecutive speeches,1 by each person, at a conference. If the order of the speakers is determined randomly, what is the probability that at least 2 of the men’s speeches will be consecutive?

A. \(\frac{3124}{3125}\)
B. \(\frac{9}{10}\)
C. \(\frac{4}{5}\)
D. \(\frac{16}{25}\)
E. \(\frac{1}{2}\)
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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each p  [#permalink]

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New post 23 Jul 2016, 00:10
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The total no of ways we can arrange 3 men and 2 women(MMMWW) is 5!/3!2!=10 as we have 3 males and 2 women.

Now there is only 1 way when two men cant be consecutive i.e. MWMWM

Therefore the prob atleast 2 men will be consecutive will be 9/10.

Though I marked the wrong answer, but after some thinking I got with the explanation.
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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each p  [#permalink]

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New post 23 Jul 2016, 11:27
so basically we have 5 seats and 5 People to sit , we can have 5! ways.= 120
lets check scenarios when no men together MWMWM ONLY POSSIBLE WAY. Men can arrange them selves in 3! and women can 2!
3*2*2/120 = 1/10
1-1/10= 9/10
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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each p  [#permalink]

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New post 23 Jul 2016, 15:04

At least scenario


\(In\; an\; at\; least\; scenario\; like\; this\; one\; the\; solution\; can\; also\; be\; found\; by\; doing\;\)

\(\frac{total\; cases\; -\; cases\; where\; there\; are\; no\; consecutive\; speeches}{total\; cases}\;\)

\(=\; \frac{5!-3\cdot 2\cdot 2\cdot 1\cdot 1}{5!} =\; \frac{108}{120}\; =\; \frac{12\cdot 9}{12\cdot 10}\; =\; \frac{9}{10}\)

B it is.

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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each p  [#permalink]

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New post 08 May 2017, 18:23
What I did was focused on the probability of non event and then subtracted this probability from 1

Say we have M1 M2 M3 W1 & W2 (total 5 persons- 3 men and 2 women)

A)Favorable event= Men give speeches consecutively
B)Non favorable event= Men do not give speeches consecutively

We try to get Probability for B)

The only type of arrangement that Men do not deliver their speeches consecutively would be as below

M W M W M

No. of ways to arrange men above would be 3! (since we have m1 m2 and m3)
No of ways to arrange women above would be 2! (since we have w1 and w2)

So total ways 3! X 2! = 12

Total no. of ways in which all five people can give speeches is 5! = 120

So Probability of B is 12/120= 1/10

Hence Probability of A is 1- 1/10= 9/10

Answer is hence B

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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each p  [#permalink]

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New post 16 May 2017, 20:34
AbdurRakib wrote:
Three men and 2 women will present 5 consecutive speeches,1 by each person, at a conference. If the order of the speakers is determined randomly, what is the probability that at least 2 of the men’s speeches will be consecutive?

A. \(\frac{3124}{3125}\)
B. \(\frac{9}{10}\)
C. \(\frac{4}{5}\)
D. \(\frac{16}{25}\)
E. \(\frac{1}{2}\)


The key word in this problem is at least - the complement rule should always register in one's mind upon seeing this term- for the sake of saving time and mastery of the concept

1- P (event not occuring) = probability the event actually does occur

1- 2c2 / 5c2 = 9/10
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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each p  [#permalink]

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New post 16 May 2017, 21:23
varundixitmro2512 wrote:
The total no of ways we can arrange 3 men and 2 women(MMMWW) is 5!/3!2!=10 as we have 3 males and 2 women.

Now there is only 1 way when two men cant be consecutive i.e. MWMWM

Therefore the prob atleast 2 men will be consecutive will be 9/10.

Though I marked the wrong answer, but after some thinking I got with the explanation.


Yes this is another plausible method w/r/t a fundamental statistical formula - # desired outcome(s)/ # total outcome (s)

We must first calculate the probability of an arrangement in which none of the boys appear consecutively or more accurately the number of arrangements in which boys do not appear consecutively over the total number of arrangements possible

Total number of arrangement possible= 5! because there are 5 elements and 5 slots

Now, the number of arrangement in which boys do not appear consecutively follows the pattern

BGBGB - though this does not actually mean the number of outcomes, or the number of occurrences, in which boys do not appear consecutively is 1 single possibility because all three boys are still distinct entities (e.g if you have three boys fred, bob, and sam then you could have (f)g(b)g(s) or (b)g(s)g(f) - etc) so the actual number of ways we can arrange the three boys with respect to that pattern is 3! More fundamentally, this question is a permutation so like wise... the number of ways we can arrange girls in that pattern is 2! . Hence

1- 3!2!/5!

Hence

9/10
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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each p  [#permalink]

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New post 28 May 2018, 18:46
AbdurRakib wrote:
Three men and 2 women will present 5 consecutive speeches,1 by each person, at a conference. If the order of the speakers is determined randomly, what is the probability that at least 2 of the men’s speeches will be consecutive?

A. \(\frac{3124}{3125}\)
B. \(\frac{9}{10}\)
C. \(\frac{4}{5}\)
D. \(\frac{16}{25}\)
E. \(\frac{1}{2}\)


Since there are 5 people, there are 5! = 120 ways the speeches can be presented. The only way the 2 of the men’s speeches will not be consecutive is if the order is (M = a man’s speech and W = a woman’s speech): M-W-M-W-M. The number of ways this can be done is 3 x 2 x 2 x 1 x 1 = 12. So there must be 120 - 12 = 108 ways that at least two of the men’s speeches will be consecutive and thus the probability is 108/120 = 9/10.

Alternate Solution:

Using the formula for permutations with indistinguishable objects, 3 men and 2 women can present speeches in 5!/(3! x 2!) = (5 x 4)/(2 x 1) = 10 ways.

Of these 10 orderings of speeches, only M-W-M-W-M does not meet the requirement of having at least two consecutive talks by men; thus, 10 - 1 = 9 orderings do meet this requirement. Thus, the probability is 9/10.

Answer: B
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Re: Three men and 2 women will present 5 consecutive speeches, 1 by each p   [#permalink] 11 Jul 2020, 06:20

Three men and 2 women will present 5 consecutive speeches, 1 by each p

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