A bags contains 5 white, 6 red and 4 blue balls. Three balls are drawn

1.	A bags contains 5 white, 6 red and 4 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red is:-1. 2/912. 4/913. 5/914. 7/91
Answer» Correct Answer - Option 2 : 4/91 Given: A bags contains 5 white, 6 red and 4 blue balls. Concept used: Factorial method used. Formula used: ⁿC_r= n!/[r! (n – r)!] Where, n = The number of items, r = How many items are taken at a time. Probability (E) = (Number of favorable outcomes)/(Total no. of possible outcomes) Calculation: Let the sample space be S. Let event of getting all the three red balls be E. The number of items = (5 + 6 + 4) ⇒ 15 According to the question: n (X) = Number of ways drawing 3 balls out of 15 ⇒ n Cr = n!/[r! (n – r)!] ⇒ ¹⁵C₃ = 15!/[3! (15 – 3)!] ⇒ 15C3 = (15!)/(3! × 12!) ⇒ 15C3 = (15 × 14 × 13 × 12!)/(3 × 2 × 1 × 12!) ⇒ 15C3 = (5 × 7 × 13) ⇒ 15C3 = 455 Again, n (Y) = ⁶C3 ⇒ 6C3 = 6!/[3! (6 – 3)!] ⇒ 6C3 = (6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1 × 3 × 2 × 1) ⇒ 6C3 = 20 Now, P (E) = n (y)/n (x) ⇒ 20/455 ⇒ 4/91 ∴ The probability that all of them are red is 4/91.

A bags contains 5 white, 6 red and 4 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red is:-1. 2/912. 4/913. 5/914. 7/91

Answer» Correct Answer - Option 2 : 4/91

Given:

A bags contains 5 white, 6 red and 4 blue balls.

Concept used:

Factorial method used.

Formula used:

ⁿC_r= n!/[r! (n – r)!]

Where, n = The number of items, r = How many items are taken at a time.

Probability (E) = (Number of favorable outcomes)/(Total no. of possible outcomes)

Calculation:

Let the sample space be S.

Let event of getting all the three red balls be E.

The number of items = (5 + 6 + 4)

⇒ 15

According to the question:

n (X) = Number of ways drawing 3 balls out of 15

⇒ n Cr = n!/[r! (n – r)!]

⇒ ¹⁵C₃ = 15!/[3! (15 – 3)!]

⇒ 15C3 = (15!)/(3! × 12!)

⇒ 15C3 = (15 × 14 × 13 × 12!)/(3 × 2 × 1 × 12!)

⇒ 15C3 = (5 × 7 × 13)

⇒ 15C3 = 455

Again, n (Y) = ⁶C3

⇒ 6C3 = 6!/[3! (6 – 3)!]

⇒ 6C3 = (6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 1 × 3 × 2 × 1)

⇒ 6C3 = 20

Now, P (E) = n (y)/n (x)

⇒ 20/455

⇒ 4/91

∴ The probability that all of them are red is 4/91.