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Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 chees

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guddo
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Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 chees [#permalink] New post   03 Jan 2024, 07:36
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Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 cheese sandwiches. Steven's picnic basket contains 3 chicken-salad sandwiches and 5 cheese sandwiches. If 1 sandwich is to be chosen at random from each basket, what is the probability that both sandwiches chosen will be chicken-salad sandwiches?

A. 1/8
B. 1/4
C. 1/3
D. 3/8
E. 1/2



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gmatophobia
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Re: Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 chees [#permalink] New post   03 Jan 2024, 13:41
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guddo wrote:
Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 cheese sandwiches. Steven's picnic basket contains 3 chicken-salad sandwiches and 5 cheese sandwiches. If 1 sandwich is to be chosen at random from each basket, what is the probability that both sandwiches chosen will be chicken-salad sandwiches?

A. 1/8
B. 1/4
C. 1/3
D. 3/8
E. 1/2



P(Cindy selecting a chicken-salad sandwich) = \(\frac{2}{6}\)

P(Steven selecting a chicken-salad sandwich) = \(\frac{3}{8}\)

P(Cindy selecting a chicken-salad sandwich AND Steven selecting a chicken-salad sandwich) = \(\frac{2}{6} * \frac{3}{8}\)

= \(\frac{1}{3} * \frac{3}{8}\)

= \(\frac{1}{8}\)

Option A
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unicornilove
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Re: Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 chees [#permalink] New post   18 Mar 2024, 22:07
Hi gmatophobia,

Why is it not necessary to multiply by 2! to consider the other case when Steven's basket is picked first?
gmatophobia wrote:
guddo wrote:
Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 cheese sandwiches. Steven's picnic basket contains 3 chicken-salad sandwiches and 5 cheese sandwiches. If 1 sandwich is to be chosen at random from each basket, what is the probability that both sandwiches chosen will be chicken-salad sandwiches?

A. 1/8
B. 1/4
C. 1/3
D. 3/8
E. 1/2


 

P(Cindy selecting a chicken-salad sandwich) = \(\frac{2}{6}\)

P(Steven selecting a chicken-salad sandwich) = \(\frac{3}{8}\)

P(Cindy selecting a chicken-salad sandwich AND Steven selecting a chicken-salad sandwich) = \(\frac{2}{6} * \frac{3}{8}\)

= \(\frac{1}{3} * \frac{3}{8}\)

= \(\frac{1}{8}\)

Option A

­
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Re: Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 chees [#permalink] New post   18 Mar 2024, 23:19
 
unicornilove wrote:
Hi gmatophobia,

Why is it not necessary to multiply by 2! to consider the other case when Steven's basket is picked first ?
­

­Hey unicornilove

As the order of selection doesn't matter do not multiply by 2!

For example, whether we choose chicken-salad sandwiches from Cindy's picnic basket or Steven's picnic basket, the outcome wouldn't matter. In the end, we need to choose from both baskets. 
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Re: Cindy's picnic basket contains 2 chicken-salad sandwiches and 4 chees [#permalink]
18 Mar 2024, 23:19
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