Question

There are 150 tickets in box numbered 1 to 150 wht is th probability of choosing a ticket which has a number a multiple of 3 or 7?

Answer 1

Answer: S=(150cards)

Let event A be the probability of choosing a ticket which has a number of multiple of 3 and 7

A=(3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99,102,105,108,111,114,117,120,123,126,129,132,135,138,141,144,147,7,14,28,35,49,56,70,77,91,98,112,119,133,140)

n(A)=63 therefore p(A)=n(A)/n(S)

Therefore 63/150=2.38~2.40

Hope it's helpful mark me as brainlist

Step-by-step explanation:

Answer 2

The probability of  choosing a ticket with a number a multiple of 3 or 7 is 32/75

Given:

There are 150 tickets in box numbered from 1 to 150

To find:

The probability of choosing a ticket with a number as multiple of 3 or 7

Solution:  

Given that the tickets numbered from 1 to 150

From 1 to 150,

Multiples of 3 = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102,105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150}

Multiples of 7 = { 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147 }  

From above Multiples of 3 or 7 from 1 to 150 = {3, 6, 7, 9, 12, 14, 15, 18, 21, 24, 27, 28, 30, 33, 35, 36, 39, 42, 45, 48, 49, 51, 54, 56, 57, 60, 63, 66, 69, 70, 72, 75, 77, 78, 81, 84, 87, 90, 91, 93, 96, 98, 99, 102, 105, 108, 111, 112, 114, 117, 119, 120, 123, 126, 129, 132, 133, 135, 138, 140, 141, 144, 147, 150}

⇒ Number of multiples of 3 or 7 = 64

Let E be the event of choosing a ticket with a number a multiple of 3 or 7  

⇒ P(E) = No. of favourable outcomes / total No. of outcomes

⇒  P(E) = 64/150 = 32/75  

The probability of  choosing a ticket with a number a multiple of 3 or 7 = 32/75

#SPJ2