Question

A box contains 10 coupons, labelled as 1,2,.....10.Three coupons are drawn at random and without  X1,X2 and X3 denote the numbers on the coupons. Then the probability that max{X1,X2,X3}<7 is 

• A. \(\dfrac{3C_1}{10C_3}\)
• B. \(\dfrac{7C_3}{10C_3}\)
• C. \(\dfrac{3C_3}{10C_3}\)
• D. \(\dfrac{3C_1}{10C_7}\)
• E. \(\dfrac{6C_3}{10C_3}\)

Answer 1

Answer: (E)

Given Data:

Total no of coupons = \(10 \) (labelled as 1,2,.....10)

Three coupons are drawn at random and without replacement.

To find  probability of max {X1,X2,X3}\(<7\)

Proceeding according to the question, drawing \(3\) coupons out of \(10\) =\(10C_3\)

Now, let's count the number of favorable outcomes (max{X1, X2, X3} < 7).

For max {X1, X2, X3} to be less than \(7\), all three coupons X1, X2, and X3 must have values less than \(7\) and there are \(6\) coupons (1 to 6) that fulfills the condition.

So, drawing \(3 \) coupons out of these \( 6\) coupons =\(6C_3 \)

Now , Probability=\( \dfrac{PE}{TE}\)  (⇢ Possible events/Total events)

                           = \(\dfrac{6C_3}{10C_3}\) (_Ans.)