Basic Probability

Question

A team of 8 couples (husband &wife) attend a Lucky draw in which 4 person picked up for a prize.then the probability that there is at least one couple is.                                                                                           (A) 11/39  (B) 12/39   (C) 14/39 (D) 15/39

Comments

D) ?

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Let me know the source of the question and whether u r also facing the same @Prateek kumar

Answer 1

There are two cases possible

case I) 1 couple + 2 individual  or case II) 2 couples

Total sample space = total ways in which 4 person can win = ¹⁶C₄ = 1820

now

case I) 1 couple + 2 individual

number of ways of choosing 1 couple = ⁸C₁ = 8

now this two individual can be both husband or both wife or different husband and wife

both husband = ⁷C₂ = 21        ( since one husband is already calculated in couple so 7 husband remains )

both wife = ⁷C₂ = 21               ( same reason )

different husband and different wife = ⁷C₁ * ⁶C₁ = 42     ( chossing 1 husband from 7 husbands and since we cant choose the wife of choosed husband. So choosing one wife from remaining 6 wives )

So total ways in this case = 8 * ( 21+21+42 ) = 672

case II) 2 couples

number of ways of choosing 2 couples = ⁸C₂ = 28

So probability = (672 + 28)/1820 = 5/13 = 15/39

So ans should be D)

Comments

I dont think that's correct @Prateek. Your numerator is greater that denominator

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@ Digvijaysingh Gautam @Habibkhan

8 pairs of shoes are in a closet. 8 shoes are selected at random. The probability that there will be at least one pair and at most 3 pairs amongst selected shoes are? 

can u help on this question too ??

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solve it similar to this. Consider the cases as

i) one pair * ( 6 left shoes + 6 right shoes + mismatch shoes )

ii) two pairs * ( 4 left shoes + 4 right shoes + mismatch shoes )

iii) three pairs * ( 2 left shoes + 2 right shoes + mismatch shoes )

Answer 2

Probability of selecting at least one couple = 1- probability of selecting no couple

Probability of selecting no couple is calculated as folows:

Let us imagine that 4 couples are standing side by side (Choosing any 4 couple from 8 couple can be done in $\binom{8}{4}$ ways). If we select exactly one person from each couple (this can be done in $2^{4}$ ways), then it is possible that no two persons who are couples are selected. So this can be done in $\binom{8}{4} * 2^{4}$ ways. Totally, 4 persons from 16 can be selected in $\binom{16}{4}$ ways. The required probability that no couple is selected is thus: $\frac{\binom{8}{4} * 2^{4}}{\binom{16}{4}} = \frac{8}{13}$

Thus, the probability that at least one couple is selected is $1-\frac{8}{13} = \frac{5}{13}$. Here the answer that matches is $\frac{15}{39}$