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Ten chocolates are distributed randomly among three children standing in a row. The probability that the first child receives exactly three chocolates is

  1. $\frac{5 \times 2^{11}}{3^9}$
  2. $\frac{5 \times 2^{10}}{3^9}$
  3. $\frac{1}{3^9}$
  4. $\frac{1}{3}$
in Probability by (37 points)
edited by | 441 views
0

If the question was like this 

Ten chocolates are distributed randomly among three children standing in a row. The probability that the first child receives  three or more chocolates is

The answer would be $(^{10}C_3 * 3^7 )∕ 3^{10}$

please correct if I am wrong

3 Answers

+12 votes
Best answer

Let the three children be $A,B$ and $C$ where $A$ is the first child

Calculating the total number of case

For each chocolate (assuming chocolates are distinguishable) we have $3$ choices, give it to $A$ or $B$ or to $C.$

So for ten chocolates we have total no. of choices $=\underbrace{3\times 3\times 3 \times \ldots \times 3}_{\text{ten times}}=3^{10}.$

Calculating the number of favorable cases

Firstly we have to select three chocolates out of ten that have to be given to $A$ and number of ways for this $={}^{10}C_3.$ 

For remaining $7$ chocolates, for each we have two choices i,e either to give it to $B$ or to $C.$

So, no of favorable cases $= {}^{10}C_3 \times 2^7.$

Our required probability $=\frac{\text{No. of favorable cases}}{\text{Total no. of cases}}$

$\quad \quad=\frac{{}^{10}C_3\times 2^7}{3^{10}} $

$\quad \quad =\frac{5\times 2^{10}}{3^9}$

i.e., option B

by Active (3.3k points)
edited by
0
did you consider chocolates distinguishable?
0
Are chocolates considered distinguishable?? Only then we need to select 3 out of 10, i.e. 10C3. I'm confused :/
0

Yes I consider them to be distinguishable because if we consider them to be undistinguished then

probability is=27/310 which is not an option.

+9

@akb1115

if you consider them indistinguishable, then the problem will be like,

probability of getting solutions of x1+x2+x3=10, where x1=3..

total outcomes = (10+3-1)C2 

favourable outcomes = (7+2-1)C1

probabity = 8C1/12C2 = 4/33

+1
@Rishabh Gupta 2

yes there is ambiguity in qsn, they should mention that chocolates are distincts..
0
@joshi_nitish your approach looks correct to me.
0

Yeah. The question should contain term distinct i.e.

Ten distinct chocolates are distributed randomly among three children standing in a row.

+9 votes

We can solve this problem using Binomial distribution. Assuming that  chocolates are distinguishable. Each chocolate has 1/3 probability to be given to the first child in the row.

P=$\frac{1}{3}$  q=1 - $\frac{1}{3}$ = $\frac{2}{3}$

10C3*$(1/3)^3 (2/3)^7$ = 5x$\frac{2^{10}}{3^9}$

by Boss (44.1k points)
edited
+2 votes
Number of ways to select 3 chocolates out of 10 = $\binom{10}{3}$ . Now these chocolates should be given to first child.

For each of the remaining 7 chocolates we have 2 choices (2 children) = $2^{7}$

Total number of ways 10 chocolates be distributed among 3 children =  $3^{10}$

So probability = $\frac{\binom{10}{3} * 2^{7}}{3^{10}} = \frac{5*2^{10}}{3^{9}}$
by Loyal (8.9k points)

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