$7$ coupons are numbered $1$ to $7.$ Four are drawn one by one with replacement. If the probability of 'any selected coupon is greater than or equal to $5^{'}$ is ${(\frac{a}{b})}^{p},$ then $a + b - p =$
(where $a, b$ are relative prime)

Open in App

Solution

$S = {1, 2, 3, 4, 5, 6, 7}$
$A = {5, 6, 7} =$ event that any selected coupon is greater than or equal to $5.$
So, $P (A) = \frac{3}{7}$
So Required probability $= P (A_{1} \cap A_{2} \cap A_{3} \cap A_{4}) =$ getting any selected coupon is greater than or equal to $5$ in all $4$ selection
$= P (A_{1}) P (A_{2}) P (A_{3}) P (A_{4}) = \frac{3}{7} \cdot \frac{3}{7} \cdot \frac{3}{7} \cdot \frac{3}{7} = {(\frac{3}{7})}^{4}$

Suggest Corrections

Similar questions

15

coupons are numbered

1, 2, 3, . . ., 15

respectively.

7

coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is

9

15 coupons are numbered 1....15 respectively.7 coupons are selected at random 1 at a time with replacement.The probability that the largest number appearing is 9?

Join BYJU'S Learning Program

7 coupons are numbered 1 to 7. Four are drawn one by one with replacement. If the probability of 'any selected coupon is greater than or equal to 5′ is (ab)p, then a+b−p= (where a,b are relative prime)

S={1,2,3,4,5,6,7} A={5,6,7}= event that any selected coupon is greater than or equal to 5. So, P(A)=37 So Required probability =P(A1∩A2∩A3∩A4)=getting any selected coupon is greater than or equal to 5 in all 4 selection =P(A1)P(A2)P(A3)P(A4)=37⋅37⋅37⋅37=(37)4

$7$ coupons are numbered $1$ to $7.$ Four are drawn one by one with replacement. If the probability of 'any selected coupon is greater than or equal to $5^{'}$ is ${(\frac{a}{b})}^{p},$ then $a + b - p =$
(where $a, b$ are relative prime)