Question

A bag contains chits numbered 5 to 35.A chit is drawn at random. What is the probability of getting a) a multiple of 3 b) a multiple of 5 c)a multiple of 3 or 5
d)a multiple of 3 and 5.

Answer 1

Step-by-step explanation:

A multiple of 5....

Hope it is correct....

Answer 2

Step-by-step explanation:

Let S be the sample space.

S = {5,6,8,9,.......,35}

n(S) = 31

Let A be the event that chit drawn is a multiple of 3

A = {6,9,12,15,18,21,24,27,30,33}

n(A) = 10

$p(a) = \frac{n(a)}{n(s)}$

Therefore,

$p(a) = \frac{10}{31}$

Therefore, probability of getting a multiple of 3 is 10/31

II)

Let B be the event that chit drawn is a multiple of 5

B = {5,10,15,20,25,30,35}

n(B) = 7

Therefore,

$p(b) = \frac{n(b)}{n(s)}$

$p(b) = \frac{7}{31}$

Therefore, probability of getting a multiple of 5 is 7/31

III)

Let C be the event that the chit drawn is multiple of 3 and 5 (in common)

C = {15,30}

n(C) = 2

therefore,

$p(c) = \frac{n(c)}{n(s)}$

i.e

$p(c) = \frac{2}{31}$

Therefore, probability of getting a multiple of 3 and 5 is 2/31

IV)

Let D be the event that chit drawn is multiple of 3 or 5 (either 3 or 5)

D = {5,6,9,10,12,15,18,20,21,24,25,27,30,33,35}

n(D) = 15

$p(d) = \frac{n(d)}{n(s)}$

therefore,

$p(a) = \frac{15}{31}$

Therefore, probability of getting a multiple of 3 or 5 is 15/31

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