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Set A contains three different positive odd integers and two

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Set A contains three different positive odd integers and two  [#permalink]

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New post 03 Aug 2014, 12:05
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Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?

(A) 6/25
(B) 2/5
(C) 1/2
(D) 3/5
(E) 19/25
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Re: Set A contains three different positive odd integers and two  [#permalink]

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New post 03 Aug 2014, 12:33
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oss198 wrote:
Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?

(A) 6/25
(B) 2/5
(C) 1/2
(D) 3/5
(E) 19/25


My approach- probability of product even(P(E)) = 1- probability of product odd(P(O))

Odd product is possible only when an integer picked from both sets is odd, so..

P(O) = (3c1* 2c1)/5c1*5c1
= 6/25
P(E) = 1-6/25 = 19/25.

Hope it helps
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Re: Set A contains three different positive odd integers and two  [#permalink]

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New post 09 Aug 2014, 01:43
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oss198 wrote:
Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?

(A) 6/25
(B) 2/5
(C) 1/2
(D) 3/5
(E) 19/25


Another approach:

P (product of 2 chosen integers is even) = P(E&E) + P(E&O) + P (O&E) = 2/5*3/5 + 2/5*2/5 + 3/5*3/5 = 19/25
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Re: Set A contains three different positive odd integers and two  [#permalink]

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New post 10 Aug 2014, 01:42
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Three approaches:
1. Backward (the shortest)
P(product even)=1-P(product odd)=1-P(odd from A)*P(odd from B)=1-3/5*2/5=19/25

2. The previous post
P (product of 2 chosen integers is even) = P(E&E) + P(E&O) + P (O&E) = 2/5*3/5 + 2/5*2/5 + 3/5*3/5 = 19/25

3. Using formula for overlapping sets
P(even from A or even from B)=P(even from A)+P(even from B)-P(even from A and even from B)=2/5+3/5-2/5*3/5=19/25
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Re: Set A contains three different positive odd integers and two  [#permalink]

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New post 20 Mar 2018, 08:24
oss198 wrote:
Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?

(A) 6/25
(B) 2/5
(C) 1/2
(D) 3/5
(E) 19/25



Actually what this question means is that all the elements are unique both within set A and also with respect to set B

Consider the following scenarios:
Scenario 1:
Set A :{ 1,3,5,2,4} Set B: {1,3,2,4,6} Here all the elements within a set are unique BUT are NOT unique with respect to the other set.
Required probability : \(\frac{3}{5}\)

Scenario 2:
Set A :{ 1,3,5,4,6} Set B: {7,9,8,10,12} Here all the elements within a set are unique AND are also unique with respect to the other set.
Required probability : \(\frac{19}{25}\)
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Re: Set A contains three different positive odd integers and two  [#permalink]

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New post 25 Mar 2018, 11:39

Solution



If the multiplication of two integers has to be even, then at least 1 integer between the two must be even.

Thus, we can calculate the probability of the product of two integers, one from set A and another from set B, to be even in two ways.

    P(A-even) * P (B- odd) + P(A-odd) * P (B- even) + P(A-even) * P (B- even), OR,
    1- P (A- odd) * P (B- odd)

Surely, the 2nd method needs very little calculation.
    • P (A- odd) = \(\frac{3}{5}\)
    • P (B- odd) = \(\frac{2}{5}\)
    • 1- P (A- odd) * P (B- odd) = 1- (\(\frac{3}{5}\)) * (\(\frac{2}{5}\)) = \(\frac{19}{25}\)

Answer: E
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Set A contains three different positive odd integers and two  [#permalink]

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New post 25 Mar 2018, 15:03
oss198 wrote:
Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?

(A) 6/25
(B) 2/5
(C) 1/2
(D) 3/5
(E) 19/25

Calculating the probability of getting an odd product seems easier.

Only one combination yields odd: odd * odd

P(odd) from set A = \(\frac{3}{5}\)

P(odd) from set B =\(\frac{2}{5}\)

\(\frac{3}{5}*\frac{2}{5}=\frac{6}{25}\)

The products are either even or odd. Probability of an even or odd product = 1, so

P(even) = 1 - P(odd)
\(\frac{25}{25}-\frac{6}{25}=\frac{19}{25}\)

Answer E
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Re: Set A contains three different positive odd integers and two  [#permalink]

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New post 05 Apr 2018, 11:04
oss198 wrote:
Set A contains three different positive odd integers and two different positive even integers; set B contains two different positive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?

(A) 6/25
(B) 2/5
(C) 1/2
(D) 3/5
(E) 19/25


We can use the formula:

1 - P(odd product) = P(even product)

We may recall that odd x odd = odd.

P(odd from set A) = 3/5

P(odd from set B) = 2/5

P(odd product) = 3/5 x 2/5 = 6/25

P(even product) = 1 - 6/25 = 19/25

Answer: E
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Re: Set A contains three different positive odd integers and two &nbs [#permalink] 05 Apr 2018, 11:04
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