Q&A: Anyone good at Data Management/Probability?
by Kaustav Bhattacharya
Question by TheDoctor: Anyone good at Data Management/Probability?
I’ve got a question I would like answered if anyone is able to.
Candice owns a chocolate shop. One of her most popular products is a box of 40 assorted chocolates, 5 which contain nuts.
a) If a person selects two chocolates at random from the box, what is the probability that:
i) Both chocolates contain nuts?
ii) Atleast one contains nuts
iii) Only one contains nuts
iv) Neither contain nuts
b) Describe how you could stimulate choosing the two chocolates. Outline a method using
i) A manual technique
ii) Appropriate technology
c) Due to the popularity of the chocolates with nuts, Candice is planning to double the number of them in each box. She claims that having 10 of the 40 chocolates contain nuts will double the popularity that one or both of two randomly selected chocolates will contain nuts. Do you agree with her claim? Support your answer with probability calculations.
Thanks!
Best answer:
Answer by Sue E
a)i) 5C2/40C2 = 10/780=1/78
ii) 1- 5C0x35C2/40C2 =1- 595/780 = 1-119/156 =37/156
iii) 5C1x35C1/40C2 = 175/780 =35/156
iv) 119/156 from ii
b) Take a pack of cards 52 cards – remove all one suit, say Clubs,except for 1 card.
ie you now have 40 cards. Treat the Ace,2,3,4,5 of hearts as the nut chocolates.
Shuffle the cards. Flip the top two.
Count the number of Ace, 2, 3 4,and 5 of hearts in these two cards.
Keep repeating the experiment lots of times.
Using technology.
Create a random number generator that generates integer numbers between 1 and 40.
Generate two numbers.
Treat the numbers 1,2,3,4,5 as the nut chocolates.
etc
c) P(0 nuts) = 10C0x30C2/40C2 = 435/780 = 29/52
ie P(at least 1 nut) = 1-29/52 = 23/52 = 0.4423
With 5 nuts, P(at least 1 nut before ) =37/156 = 0.2371
So she is almost on the money – but not quite – so her claim is incorrect.
Add your own answer in the comments!
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about 1 year ago
The probability that a randomly selected chocolate contains nuts is 5/40 or .1250.
i) The probability that both randomly selected chocolates contain nuts is ( 5/40 ) ( 4/39 ) = .01282 because the first one contains nuts, and you don’t put it back (therefore the trials are not independent), so there are only four chocolates with nuts in the box of thirty-nine.
ii) The probability that “at least one contains nuts” is the probability that one or two contains nuts. This will be the probability that one contains nuts plus the probability that both contain nuts:
(35/40) (5/39) + (5/40) (35/39) + (5/40) (4/39) = .2372
where the first product is the probability that the first chocolate does not have nuts and the second one does, the second product is the probability that the first one does and the second doesn’t, and the third product is the probability that the first does and the second does.
iii) The probability that only one contains nuts is the sum of the first two products from above:
(35/40) (5/39) + (5/40) (35/39) = .2244
iv) The probability that neither contain nuts is the probability that the first one doesn’t and the second one doesn’t
(35/40) (34/39) = .7628
Notice this is the complement of part i) where both contained nuts.
b) To simulate this experiment, I would go buy a box of 40 chocolates and stuff a nut into five of them and then shake the box. Then I would randomly select two and eat them. Seriously, it’s kind of a dumb question. I know they’re looking for “roll two forty-sided dice and the numbers 1 – 5 represent chocolates with nuts” but eating chocolate is so much better!
c) The probability that one of the two randomly selected chocolates will contain nuts in this new configuration is:
(10/40) (30/39) + (30/40) (10/39) = .3846
and the probability that they both contain nuts is:
(10/40) (9/39) = .0577
Their sum (.4423) is the probability that one or both contain nuts, or at least one contains nuts.
Notice none of these are double their counterparts in the original configuration.
*wowee* that was a long one, man!