Test: Probability- 5 - Bank Exams MCQ

(2/3) x (3/4) + (1/3) x (2/7) = (1/2) + (2/21) = (25/42)

They will contradict each other if: A is true and B is false or A is false and B is true.

(3/4) x (1/6) + (1/4) x (5/6) = 1/3.

The whole numbers selected can only be 1, 3, 7 or 9 and cannot contain 2, 4, 6, 8, 0 or 5.

11/17 (if the first one is black, there will be 11 black balls left out of 17)

There will be ⁶C₃ triangles formed overall. Out of these visualise the number of equilateral triangles.

We do not have to consider any sum other than 5 or 7 occurring.
A sum of 5 can be obtained by any of [4 + 1, 3 + 2, 2 + 3, 1 + 4]
Similarly a sum of 7 can be obtained by any of [6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6]
For 6: n(E) = 4, n(S) = 6 + 4 P = 0.4
For 7: n(E) = 6 n(S) = 6 + 4 P = 0.6

This problem has to be treated as if we are selecting the third card out of the 50 remaining cards. 11 of these are spades.
Hence, 11/50.

A total o f 6 can be obtained by either (1 + 2 + 3) or by (2 + 2 + 2).

Try to find the number of ways in which 0 or 1 bomb hits the bridge if n bombs are thrown.
The required value of the number of bombs will be such that the probability of 0 or 1 bomb hitting the bridge should be less than 0.1.

Explanation: We have first 50 natural numbers.

There are four common multiples of 3 and 4 from the first 50 natural numbers that are = 12, 24, 36, 48

So, P(multiple of 3 and 4) = 4/50 or 2/25

Hence, the correct option is 'A'.

In the first draw, we have 7 even tickets out of 15 and in the second we have 8 odd tickets out of 15.
Thus, (7/15) x (8/15) = 56/225.

Event definition: Any six of them work AND four leave OR any seven work AND three leave OR any eight work AND two leave OR any nine work AND one leaves OR All ten work.

There are three ways draw 2 white balls and 1 black ball from the three boxes. They are:

Scenario 1: white ball from box 1, white ball from box 2, black ball from box 3

Scenario 2: white ball from box 1, black ball from box 2, white ball from box 3

Scenario 3: black ball from box 1, white ball from box 2, white ball from box 3

For box #1: P(white ball) = 3/4 = 0.75 and P(black ball) = 1/4 = 0.25

For box #2: P(white ball) = 2/4 = 0.5 and P(black ball)= 2/4 = 0.5

For box #3: P(white ball) = 1/4 = 0.25 and P(black ball) = 3/4 = 0.75

The probability of scenario 1 occurring is 0.75 * 0.5 * 0.75 = 0.28125

The probability of scenario 2 occurring is 0.75 * 0.5 * 0.25 = 0.09375

The probability of scenario 3 occurring is 0.25 * 0.5 * 0.25 = 0.03125

Thus, the probability that 2 white balls and 1 black ball will be drawn is P(scenario 1) + P(scenario 2) + P(scenario 3) = 0.28125 + 0.09375 + 0.03125 = 0.40625 = 13/32