A coin is tossed 7 times. The number of times head comes is 5, so the probability of getting tail is

(a.) $ \frac{2}{7} $

(b.) $ \frac{5}{7} $
(c.) $ \frac{1}{7} $
(d.) $ \frac{4}{7} $

Given:

A coin is tossed 7 times.  The number of times head comes is 5.

To do:

We have to find the probability of getting a tail.

Solution:

The total number of outcomes$=7$.

Number of outcomes in which head comes$=5$

This implies,

Number of outcomes in which tail comes$=7-5=2$

We know that,

Probability of an event=$ \frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes}$

Therefore,

Probability of getting tail$=\frac{2}{7}$.

The probability of getting tail is $\frac{2}{7}$.

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Updated on: 10-Oct-2022

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