Question

Tickets numbered from 1 to 40 are in a bag on ticket is drawn at random what is the probability that the card drawn is a multiple of 3 or 7

Answer 1

Answer:

17/40

Step-by-step explanation:

NO OF MULTIPLES OF 3 IN 1-40= 13

NO OF MULTIPLES OF 7 IN 40 = 5

BUT 21 IS A MULTIPLE OF BOTH 3 AND 7. SO IT WILL BE COUNTED ONLY ONCE

SO TOTAL NO OF FAVOURABLE OUTCOME= 13+4 = 17

P(MULTIPLE OF 3 OR 7 ) = FAV OTCOMES / TOTAL NO OF OUTCOMES

= 17/40

THIS IS THE PROBABILITY

PLS MARK AS BRAINLIEST AND THANKS

Answer 2

Answer:

the probability that the card drawn is a multiple of 3 or 7 is $\frac{3}{8}$

Step-by-step explanation:

Given: Tickets numbered from 1 to 40 are in a bag on ticket is drawn at random

To find: the probability that the card drawn is a multiple of 3 or 7

Solution:

Find the number of favorable and total outcomes.

Total no. of outcomes $=40$

No. of favorable outcomes which are multiples of $3=12=$$(3,6,9,12,15,18,21,24,27,30,33,36)$

$\because Probability =\frac{\text { Favorable outcomes }}{\text { Total outcomes }}$

Probability $=\frac{12}{40}=\frac{3}{10}$

The probability that the card drawn is a multiple of 3 is $\frac{3}{10}$

Find the number of favorable and total outcomes.

Total no. of outcomes $=40$

No. of favorable outcomes which are multiples of $7=5=$$(7,14,21,28,35)$

$\because Probability =\frac{\text { Favorable outcomes }}{\text { Total outcomes }}$

Probability $=\frac{5}{40}=\frac{1}{8}$

Probability that the card drawn is a multiple of 7 is $\frac{1}{8}$

21 are common in both 3 and 5, so we choose them only once. 12+5-2= 15

Probability = $\frac{15}{40} \\\frac{3}{8}$