There are 9 Black pen, 5 red pen and 3 blue pen if 2 pen are selected

1.	There are 9 Black pen, 5 red pen and 3 blue pen if 2 pen are selected blindly, find probability of getting both are red pen or blue pen.1. 13/1362. 13/3043. 11/3384. 15/288
Answer» Correct Answer - Option 1 : 13/136 Given: Black pen = 9 Red pen = 5 Blue pen = 3 Formula used: ⁿC_r = n!/((n-r)! × r!) Calculation: Let S be the sample space then, n(S) = number of way of selecting 2 pens out of 17 pens ⇒ n(S) = ¹⁷C₂ = 17!/((17– 2)! × 2!) ⇒ (17 × 16)/2 ⇒ 136 Let E be the event of getting both are red or blue pen n(E) = ⁵C₂+³C₂ ⇒ (5 × 4)/2 + (3 × 2)/2 ⇒ 10 + 3 ⇒ 13 P(E) = n(E)/n(S) = 13/136 ∴ The probability of getting both are red pen or blue pen is 13/136.

There are 9 Black pen, 5 red pen and 3 blue pen if 2 pen are selected blindly, find probability of getting both are red pen or blue pen.1. 13/1362. 13/3043. 11/3384. 15/288

Answer» Correct Answer - Option 1 : 13/136

Given:

Black pen = 9

Red pen = 5

Blue pen = 3

Formula used:

ⁿC_r = n!/((n-r)! × r!)

Calculation:

Let S be the sample space then,

n(S) = number of way of selecting 2 pens out of 17 pens

⇒ n(S) = ¹⁷C₂ = 17!/((17– 2)! × 2!)

⇒ (17 × 16)/2

⇒ 136

Let E be the event of getting both are red or blue pen

n(E) = ⁵C₂+³C₂

⇒ (5 × 4)/2 + (3 × 2)/2

⇒ 10 + 3

⇒ 13

P(E) = n(E)/n(S)

= 13/136

∴ The probability of getting both are red pen or blue pen is 13/136.

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There are 9 Black pen, 5 red pen and 3 blue pen if 2 pen are selected blindly, find probability of getting both are red pen or blue pen.1. 13/1362. 13/3043. 11/3384. 15/288

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